{"id":4351,"date":"2026-08-18T12:25:51","date_gmt":"2026-08-18T12:25:51","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=4351"},"modified":"2026-08-18T14:25:26","modified_gmt":"2026-08-18T14:25:26","slug":"integration-of-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/integration-of-trigonometric-functions.html","title":{"rendered":"Integration of Trigonometric Functions: Basic Methods and Transformations"},"content":{"rendered":"

Integration of trigonometric functions does not always come down to directly using a table of antiderivatives<\/a>. Often, the integrand first needs to be simplified using trigonometric identities or a substitution. How can you determine which method to use? To do this, you should analyze the types of functions involved, their powers, and the relationship between the individual factors.<\/p>\n

Basic Antiderivatives: Standard Formulas and Linear Arguments<\/h2>\n

The simplest trigonometric integrals can be evaluated directly using a table of antiderivatives<\/a>:<\/p>\n

\\[
\n\\begin{gathered}
\n\\int \\sin(x)\\,dx=-\\cos(x)+C,\\\\[4pt]
\n\\int \\cos(x)\\,dx=\\sin(x)+C,\\\\[4pt]
\n\\int \\frac{dx}{\\cos^2(x)}=\\tan(x)+C,\\\\[4pt]
\n\\int \\frac{dx}{\\sin^2(x)}=-\\cot(x)+C.
\n\\end{gathered}
\n\\]<\/p>\n

For tangent and cotangent, we use the formulas<\/p>\n

\\[
\n\\begin{gathered}
\n\\int \\tan(x)\\,dx=-\\ln|\\cos(x)|+C,\\\\[4pt]
\n\\int \\cot(x)\\,dx=\\ln|\\sin(x)|+C.
\n\\end{gathered}
\n\\]<\/p>\n

Why does a logarithm appear in the last two formulas? Tangent and cotangent can be written as ratios of sine and cosine:<\/p>\n

\\[
\n\\tan(x)=\\frac{\\sin(x)}{\\cos(x)},
\n\\qquad
\n\\cot(x)=\\frac{\\cos(x)}{\\sin(x)}.
\n\\]<\/p>\n

Therefore, the corresponding integrals have the structure of the standard expression<\/p>\n

\\[
\n\\int \\frac{f'(x)}{f(x)}\\,dx=\\ln|f(x)|+C.
\n\\]<\/p>\n

For example, the derivative of \\( \\cos(x) \\) is \\( -\\sin(x) \\). This is why a minus sign appears before the logarithm when integrating the tangent function.<\/p>\n

Special attention should be given to trigonometric functions with a linear argument. If \\( a\\neq 0 \\), then<\/p>\n

\\[
\n\\begin{gathered}
\n\\int \\sin(a\\cdot x+b)\\,dx
\n=
\n-\\frac{1}{a}\\cdot\\cos(a\\cdot x+b)+C,\\\\[4pt]
\n\\int \\cos(a\\cdot x+b)\\,dx
\n=
\n\\frac{1}{a}\\cdot\\sin(a\\cdot x+b)+C.
\n\\end{gathered}
\n\\]<\/p>\n

The coefficient of \\( x \\) appears in the denominator. Why does this happen? When differentiating a composite function, an additional factor \\( a \\) appears. Therefore, during integration, this factor must be compensated for.<\/p>\n

The same principle applies to other trigonometric functions:<\/p>\n

\\[
\n\\begin{gathered}
\n\\int \\frac{dx}{\\cos^2(a\\cdot x+b)}
\n=
\n\\frac{1}{a}\\cdot\\tan(a\\cdot x+b)+C,\\\\[4pt]
\n\\int \\frac{dx}{\\sin^2(a\\cdot x+b)}
\n=
\n-\\frac{1}{a}\\cdot\\cot(a\\cdot x+b)+C.
\n\\end{gathered}
\n\\]<\/p>\n

Thus, before applying a standard integration formula, you need to check not only the type of trigonometric function but also its argument.<\/p>\n

Integration of Trigonometric Functions: Powers of Sine and Cosine<\/h2>\n

After the basic formulas, let us move on to integrals that contain powers of sine and cosine:<\/p>\n

\\[
\n\\int \\sin^m(x)\\cdot\\cos^n(x)\\,dx.
\n\\]<\/p>\n

Which transformation should we choose here? First, we need to determine whether the exponents \\( m \\) and \\( n \\) are even or odd. This determines which function is most convenient to choose as the new variable.<\/p>\n

If the power of sine is odd, one factor of \\( \\sin(x) \\) is separated. The remaining power is transformed using the identity<\/p>\n

\\[
\n\\sin^2(x)=1-\\cos^2(x).
\n\\]<\/p>\n

As a result, all even powers of sine are expressed in terms of cosine. After that, we use the substitution<\/p>\n

\\[
\nu=\\cos(x),
\n\\qquad
\ndu=-\\sin(x)\\,dx.
\n\\]<\/p>\n

Let us look at the structure of this transformation:<\/p>\n

\\[
\n\\int \\sin^3(x)\\cdot\\cos^2(x)\\,dx.
\n\\]<\/p>\n

We write the odd power of sine as<\/p>\n

\\[
\n\\sin^3(x)=\\sin(x)\\cdot\\sin^2(x).
\n\\]<\/p>\n

Next, we replace the square of sine:<\/p>\n

\\[
\n\\sin^3(x)
\n=
\n\\sin(x)\\cdot\\bigl(1-\\cos^2(x)\\bigr).
\n\\]<\/p>\n

Therefore, the original integral becomes<\/p>\n

\\[
\n\\int
\n\\sin(x)\\cdot
\n\\bigl(1-\\cos^2(x)\\bigr)\\cdot
\n\\cos^2(x)\\,dx.
\n\\]<\/p>\n

After the substitution \\( u=\\cos(x) \\), we obtain an algebraic integral:<\/p>\n

\\[
\n-\\int \\bigl(1-u^2\\bigr)\\cdot u^2\\,du.
\n\\]<\/p>\n

Thus, the separated factor \\( \\sin(x) \\), together with \\( dx \\), forms the differential of the new variable.<\/p>\n

If the power of cosine is odd, we proceed in a similar way. One factor of \\( \\cos(x) \\) is separated, while the remaining power is transformed using the identity<\/p>\n

\\[
\n\\cos^2(x)=1-\\sin^2(x).
\n\\]<\/p>\n

Then we use the substitution<\/p>\n

\\[
\nu=\\sin(x),
\n\\qquad
\ndu=\\cos(x)\\,dx.
\n\\]<\/p>\n

For example, the power of cosine can be transformed as follows:<\/p>\n

\\[
\n\\cos^3(x)
\n=
\n\\cos(x)\\cdot\\cos^2(x)
\n=
\n\\cos(x)\\cdot\\bigl(1-\\sin^2(x)\\bigr).
\n\\]<\/p>\n

After that, the entire integrand can be expressed in terms of \\( \\sin(x) \\), while the factor \\( \\cos(x)\\,dx \\) can be replaced by \\( du \\).<\/p>\n

What should we do when both exponents are even? In this case, we cannot separate a single sine or cosine factor that would immediately form the differential of a new variable. Therefore, we use the power-reduction formulas:<\/p>\n

\\[
\n\\begin{gathered}
\n\\sin^2(x)=\\frac{1-\\cos(2\\cdot x)}{2},\\\\[4pt]
\n\\cos^2(x)=\\frac{1+\\cos(2\\cdot x)}{2}.
\n\\end{gathered}
\n\\]<\/p>\n

For example,<\/p>\n

\\[
\n\\sin^2(x)\\cdot\\cos^2(x)
\n=
\n\\frac{1-\\cos(2\\cdot x)}{2}
\n\\cdot
\n\\frac{1+\\cos(2\\cdot x)}{2}.
\n\\]<\/p>\n

Next, multiply the resulting expressions and simplify the result. As a result, the powers of the trigonometric functions are reduced. If a squared trigonometric function appears again, apply the power-reduction formula once more.<\/p>\n

Thus, when one of the powers is odd, we usually separate one factor and use a substitution. If both powers are even, we reduce them step by step using the power-reduction formulas.<\/p>\n

Integration of Trigonometric Functions: Powers of Tangent, Cotangent, Secant, and Cosecant<\/h2>\n

Now let us consider integrals that contain tangent and secant. Secant is defined as the reciprocal of cosine:<\/p>\n

\\[
\n\\sec(x)=\\frac{1}{\\cos(x)}.
\n\\]<\/p>\n

Cosecant is defined as the reciprocal of sine:<\/p>\n

\\[
\n\\csc(x)=\\frac{1}{\\sin(x)}.
\n\\]<\/p>\n

Integrals involving powers of tangent and secant usually have the form<\/p>\n

\\[
\n\\int \\tan^m(x)\\cdot\\sec^n(x)\\,dx.
\n\\]<\/p>\n

To transform them, we use the identities<\/p>\n

\\[
\n\\begin{gathered}
\n\\sec^2(x)=1+\\tan^2(x),\\\\[4pt]
\n\\tan^2(x)=\\sec^2(x)-1.
\n\\end{gathered}
\n\\]<\/p>\n

The following derivatives are also important:<\/p>\n

\\[
\n\\begin{gathered}
\n\\bigl(\\tan(x)\\bigr)’=\\sec^2(x),\\\\[4pt]
\n\\bigl(\\sec(x)\\bigr)’=\\sec(x)\\cdot\\tan(x).
\n\\end{gathered}
\n\\]<\/p>\n

These relationships determine how we choose the substitution.<\/p>\n

First, we check the power of secant. If it is even, we separate one factor \\( \\sec^2(x) \\). The remaining even power of secant is expressed in terms of tangent, and then we use the substitution<\/p>\n

\\[
\nu=\\tan(x),
\n\\qquad
\ndu=\\sec^2(x)\\,dx.
\n\\]<\/p>\n

Let us look at an intermediate transformation:<\/p>\n

\\[
\n\\int \\tan^3(x)\\cdot\\sec^4(x)\\,dx.
\n\\]<\/p>\n

The power of secant is even, so we write it as<\/p>\n

\\[
\n\\sec^4(x)=\\sec^2(x)\\cdot\\sec^2(x).
\n\\]<\/p>\n

We replace one of the factors using the identity<\/p>\n

\\[
\n\\sec^2(x)=1+\\tan^2(x).
\n\\]<\/p>\n

Then we obtain<\/p>\n

\\[
\n\\int
\n\\tan^3(x)\\cdot
\n\\bigl(1+\\tan^2(x)\\bigr)\\cdot
\n\\sec^2(x)\\,dx.
\n\\]<\/p>\n

After the substitution \\( u=\\tan(x) \\), the integral takes an algebraic form:<\/p>\n

\\[
\n\\int u^3\\cdot\\bigl(1+u^2\\bigr)\\,du.
\n\\]<\/p>\n

If the power of secant is odd, we check the power of tangent. When the power of tangent is also odd, we separate the product<\/p>\n

\\[
\n\\sec(x)\\cdot\\tan(x).
\n\\]<\/p>\n

The remaining even power of tangent is expressed in terms of secant using the formula<\/p>\n

\\[
\n\\tan^2(x)=\\sec^2(x)-1.
\n\\]<\/p>\n

After that, we use the substitution<\/p>\n

\\[
\nu=\\sec(x),
\n\\qquad
\ndu=\\sec(x)\\cdot\\tan(x)\\,dx.
\n\\]<\/p>\n

For example, an odd power of tangent can be written as<\/p>\n

\\[
\n\\tan^3(x)
\n=
\n\\tan^2(x)\\cdot\\tan(x)
\n=
\n\\bigl(\\sec^2(x)-1\\bigr)\\cdot\\tan(x).
\n\\]<\/p>\n

If there is also a factor \\( \\sec(x) \\), then the product \\( \\sec(x)\\cdot\\tan(x)\\,dx \\) forms the differential of the new variable.<\/p>\n

If neither of these rules provides a convenient transformation, another trigonometric identity or integration by parts may be needed.<\/p>\n

A similar approach is used for powers of cotangent and cosecant. Here we use the identity<\/p>\n

\\[
\n\\csc^2(x)=1+\\cot^2(x)
\n\\]<\/p>\n

and the derivatives<\/p>\n

\\[
\n\\begin{gathered}
\n\\bigl(\\cot(x)\\bigr)’=-\\csc^2(x),\\\\[4pt]
\n\\bigl(\\csc(x)\\bigr)’=-\\csc(x)\\cdot\\cot(x).
\n\\end{gathered}
\n\\]<\/p>\n

A minus sign appears in these formulas. Therefore, it must always be taken into account when performing the substitution.<\/p>\n

Products of Trigonometric Functions: Converting Products into Sums<\/h2>\n

In the previous sections, we considered powers of trigonometric functions with the same argument. Now let us move on to products of sine and cosine functions whose arguments may be different.<\/p>\n

Integrating such products directly is usually inconvenient. How can we change their structure? For this purpose, we use the product-to-sum formulas:<\/p>\n

\\[
\n\\begin{gathered}
\n\\sin(\\alpha)\\cdot\\sin(\\beta)
\n=
\n\\frac{1}{2}\\cdot
\n\\bigl(
\n\\cos(\\alpha-\\beta)-\\cos(\\alpha+\\beta)
\n\\bigr),\\\\[4pt]
\n\\cos(\\alpha)\\cdot\\cos(\\beta)
\n=
\n\\frac{1}{2}\\cdot
\n\\bigl(
\n\\cos(\\alpha-\\beta)+\\cos(\\alpha+\\beta)
\n\\bigr),\\\\[4pt]
\n\\sin(\\alpha)\\cdot\\cos(\\beta)
\n=
\n\\frac{1}{2}\\cdot
\n\\bigl(
\n\\sin(\\alpha+\\beta)+\\sin(\\alpha-\\beta)
\n\\bigr).
\n\\end{gathered}
\n\\]<\/p>\n

For example, for the integral<\/p>\n

\\[
\n\\int \\sin(3\\cdot x)\\cdot\\cos(2\\cdot x)\\,dx
\n\\]<\/p>\n

we use the third formula:<\/p>\n

\\[
\n\\sin(3\\cdot x)\\cdot\\cos(2\\cdot x)
\n=
\n\\frac{1}{2}\\cdot
\n\\bigl(
\n\\sin(5\\cdot x)+\\sin(x)
\n\\bigr).
\n\\]<\/p>\n

Therefore, the original integral becomes<\/p>\n

\\[
\n\\frac{1}{2}\\cdot
\n\\int
\n\\bigl(
\n\\sin(5\\cdot x)+\\sin(x)
\n\\bigr)\\,dx.
\n\\]<\/p>\n

The resulting expression no longer contains a product. It is now a sum of standard trigonometric functions with linear arguments.<\/p>\n

It is also useful to remember the formula<\/p>\n

\\[
\n\\sin(2\\cdot x)=2\\cdot\\sin(x)\\cdot\\cos(x).
\n\\]<\/p>\n

It allows us to replace the product \\( \\sin(x)\\cdot\\cos(x) \\) with a single trigonometric function:<\/p>\n

\\[
\n\\sin(x)\\cdot\\cos(x)
\n=
\n\\frac{1}{2}\\cdot\\sin(2\\cdot x).
\n\\]<\/p>\n

Thus, for products of functions with different arguments, we use the product-to-sum formulas. If the arguments are the same, the double-angle formula for sine may sometimes be enough.<\/p>\n

Integration of Trigonometric Functions: Rational Expressions with Sine and Cosine<\/h2>\n

More challenging integrals have the form<\/p>\n

\\[
\n\\int R\\bigl(\\sin(x),\\cos(x)\\bigr)\\,dx,
\n\\]<\/p>\n

where \\( R \\) is a rational function of \\( \\sin(x) \\) and \\( \\cos(x) \\). This means that sine and cosine appear in the expression through addition, subtraction, multiplication, division, and integer powers.<\/p>\n

First, we should check whether the integrand contains a function together with its derivative, or an expression proportional to that derivative. In this case, a simple substitution is enough.<\/p>\n

If the expression contains \\( \\sin(x) \\) together with the factor \\( \\cos(x)\\,dx \\), it is convenient to set<\/p>\n

\\[
\nu=\\sin(x),
\n\\qquad
\ndu=\\cos(x)\\,dx.
\n\\]<\/p>\n

For example, in the integral<\/p>\n

\\[
\n\\int \\frac{\\cos(x)}{1+\\sin(x)}\\,dx
\n\\]<\/p>\n

the numerator is the derivative of sine. Therefore, after the substitution \\( u=\\sin(x) \\), we obtain<\/p>\n

\\[
\n\\int \\frac{du}{1+u}.
\n\\]<\/p>\n

The trigonometric functions disappear, and the original integral is reduced to an integral of a rational function of the variable \\( u \\).<\/p>\n

If the integrand contains \\( \\cos(x) \\) together with the factor \\( \\sin(x)\\,dx \\), it is convenient to use the substitution<\/p>\n

\\[
\nu=\\cos(x),
\n\\qquad
\ndu=-\\sin(x)\\,dx.
\n\\]<\/p>\n

For example,<\/p>\n

\\[
\n\\int \\frac{\\sin(x)}{2+\\cos(x)}\\,dx
\n\\]<\/p>\n

after the substitution \\( u=\\cos(x) \\) becomes<\/p>\n

\\[
\n-\\int \\frac{du}{2+u}.
\n\\]<\/p>\n

The minus sign appears because of the derivative of cosine. Therefore, it must be preserved when changing to the new variable.<\/p>\n

Sometimes the integrand contains the derivative of tangent. In that case, we use the substitution<\/p>\n

\\[
\nu=\\tan(x),
\n\\qquad
\ndu=\\frac{dx}{\\cos^2(x)}.
\n\\]<\/p>\n

For example,<\/p>\n

\\[
\n\\int
\n\\frac{dx}
\n{\\cos^2(x)\\cdot\\bigl(1+\\tan(x)\\bigr)}
\n\\]<\/p>\n

after the substitution \\( u=\\tan(x) \\) becomes<\/p>\n

\\[
\n\\int \\frac{du}{1+u}.
\n\\]<\/p>\n

This substitution is convenient because the factor \\( 1\/\\cos^2(x) \\) is exactly the derivative of tangent.<\/p>\n

However, a simple substitution does not always work. If both sine and cosine appear in the numerator or denominator and the derivative of neither function provides the required factor, we use the universal trigonometric substitution<\/p>\n

\\[
\nt=\\tan\\left(\\frac{x}{2}\\right).
\n\\]<\/p>\n

With this substitution, sine and cosine can be expressed as rational functions of \\( t \\):<\/p>\n

\\[
\n\\begin{gathered}
\n\\sin(x)=\\frac{2\\cdot t}{1+t^2},\\\\[4pt]
\n\\cos(x)=\\frac{1-t^2}{1+t^2},\\\\[4pt]
\ndx=\\frac{2}{1+t^2}\\,dt.
\n\\end{gathered}
\n\\]<\/p>\n

Why is this substitution called universal? It allows us to eliminate both sine and cosine from the integrand at the same time.<\/p>\n

Consider the structure of the integral<\/p>\n

\\[
\n\\int \\frac{dx}{2+\\sin(x)+\\cos(x)}.
\n\\]<\/p>\n

After the substitution \\( t=\\tan(x\/2) \\), we obtain<\/p>\n

\\[
\n\\int
\n\\frac{
\n\\frac{2}{1+t^2}\\,dt
\n}{
\n2+
\n\\frac{2\\cdot t}{1+t^2}
\n+
\n\\frac{1-t^2}{1+t^2}
\n}.
\n\\]<\/p>\n

Bringing the expressions in the denominator to a common denominator gives<\/p>\n

\\[
\n\\int
\n\\frac{2}{t^2+2\\cdot t+3}\\,dt.
\n\\]<\/p>\n

Thus, all trigonometric functions disappear, and the original integral is transformed into an integral of a rational function.<\/p>\n

The universal substitution works for a wide range of expressions. However, it often leads to longer algebraic transformations. Therefore, before using it, we should check whether a simpler substitution involving \\( \\sin(x) \\), \\( \\cos(x) \\), or \\( \\tan(x) \\) can be applied.<\/p>\n

Integration of Trigonometric Functions: Step-by-Step Solutions to Practical Problems<\/h2>\n

Let us see how the main integration methods are applied to different trigonometric expressions. In each case, we will first analyze the structure of the integrand and then choose the appropriate identity or substitution.<\/p>\n

Example 1. Evaluate the integral
\n\\[
\n\\int \\sin^2(x)\\cdot\\cos^3(x)\\,dx.
\n\\]<\/h3>\n

The integrand contains an odd power of cosine. Therefore, we separate one factor \\( \\cos(x) \\) and express the square of cosine in terms of sine:<\/p>\n

\\[
\n\\cos^3(x)
\n=
\n\\cos^2(x)\\cdot\\cos(x).
\n\\]<\/p>\n

We use the identity<\/p>\n

\\[
\n\\cos^2(x)=1-\\sin^2(x).
\n\\]<\/p>\n

Then<\/p>\n

\\[
\n\\cos^3(x)
\n=
\n\\bigl(1-\\sin^2(x)\\bigr)\\cdot\\cos(x).
\n\\]<\/p>\n

The original integral becomes<\/p>\n

\\[
\n\\int
\n\\sin^2(x)\\cdot
\n\\bigl(1-\\sin^2(x)\\bigr)\\cdot
\n\\cos(x)\\,dx.
\n\\]<\/p>\n

Now we make the substitution<\/p>\n

\\[
\nu=\\sin(x).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\ndu=\\cos(x)\\,dx,
\n\\]<\/p>\n

after the substitution we obtain<\/p>\n

\\[
\n\\int
\nu^2\\cdot\\bigl(1-u^2\\bigr)\\,du.
\n\\]<\/p>\n

Expand the expression:<\/p>\n

\\[
\n\\int \\bigl(u^2-u^4\\bigr)\\,du.
\n\\]<\/p>\n

Integrate each term:<\/p>\n

\\[
\n\\int \\bigl(u^2-u^4\\bigr)\\,du
\n=
\n\\frac{u^3}{3}
\n–
\n\\frac{u^5}{5}
\n+
\nC.
\n\\]<\/p>\n

Now return to the variable \\( x \\):<\/p>\n

\\[
\nu=\\sin(x).
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int \\sin^2(x)\\cdot\\cos^3(x)\\,dx
\n=
\n\\frac{\\sin^3(x)}{3}
\n–
\n\\frac{\\sin^5(x)}{5}
\n+
\nC.
\n\\]<\/p>\n

Example 2. Evaluate the integral
\n\\[
\n\\int \\sin^2(x)\\cdot\\cos^2(x)\\,dx.
\n\\]<\/h3>\n

Both trigonometric functions have even powers. Therefore, we cannot separate a single factor for a convenient substitution. In this case, we use power-reduction formulas.<\/p>\n

First, we use the double-angle formula for sine:<\/p>\n

\\[
\n\\sin(2\\cdot x)
\n=
\n2\\cdot\\sin(x)\\cdot\\cos(x).
\n\\]<\/p>\n

Hence,<\/p>\n

\\[
\n\\sin(x)\\cdot\\cos(x)
\n=
\n\\frac{1}{2}\\cdot\\sin(2\\cdot x).
\n\\]<\/p>\n

Square both sides:<\/p>\n

\\[
\n\\sin^2(x)\\cdot\\cos^2(x)
\n=
\n\\frac{1}{4}\\cdot\\sin^2(2\\cdot x).
\n\\]<\/p>\n

Therefore, the original integral becomes<\/p>\n

\\[
\n\\frac{1}{4}\\cdot
\n\\int \\sin^2(2\\cdot x)\\,dx.
\n\\]<\/p>\n

Now apply the power-reduction formula:<\/p>\n

\\[
\n\\sin^2(2\\cdot x)
\n=
\n\\frac{1-\\cos(4\\cdot x)}{2}.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\frac{1}{4}\\cdot
\n\\int \\sin^2(2\\cdot x)\\,dx
\n=
\n\\frac{1}{8}\\cdot
\n\\int
\n\\bigl(1-\\cos(4\\cdot x)\\bigr),dx.
\n\\]<\/p>\n

Split it into two integrals:<\/p>\n

\\[
\n\\frac{1}{8}\\cdot\\int dx
\n–
\n\\frac{1}{8}\\cdot\\int\\cos(4\\cdot x)\\,dx.
\n\\]<\/p>\n

The first integral is<\/p>\n

\\[
\n\\int dx=x.
\n\\]<\/p>\n

For the second integral, we take the linear argument into account:<\/p>\n

\\[
\n\\int\\cos(4\\cdot x)\\,dx
\n=
\n\\frac{1}{4}\\cdot\\sin(4\\cdot x).
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int \\sin^2(x)\\cdot\\cos^2(x)\\,dx
\n=
\n\\frac{x}{8}
\n–
\n\\frac{1}{32}\\cdot\\sin(4\\cdot x)
\n+
\nC.
\n\\]<\/p>\n

Example 3. Evaluate the integral
\n\\[
\n\\int \\tan^2(x)\\cdot\\sec^4(x)\\,dx.
\n\\]<\/h3>\n

The power of secant is even. Therefore, we separate one factor \\( \\sec^2(x) \\), which together with \\( dx \\) will form the differential of tangent:<\/p>\n

\\[
\n\\sec^4(x)
\n=
\n\\sec^2(x)\\cdot\\sec^2(x).
\n\\]<\/p>\n

Rewrite the original integral as<\/p>\n

\\[
\n\\int
\n\\tan^2(x)\\cdot
\n\\sec^2(x)\\cdot
\n\\sec^2(x)\\,dx.
\n\\]<\/p>\n

We express one of the factors \\( \\sec^2(x) \\) in terms of tangent:<\/p>\n

\\[
\n\\sec^2(x)=1+\\tan^2(x).
\n\\]<\/p>\n

Then we obtain<\/p>\n

\\[
\n\\int
\n\\tan^2(x)\\cdot
\n\\bigl(1+\\tan^2(x)\\bigr)\\cdot
\n\\sec^2(x)\\,dx.
\n\\]<\/p>\n

Now make the substitution<\/p>\n

\\[
\nu=\\tan(x).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\ndu=\\sec^2(x)\\,dx,
\n\\]<\/p>\n

the integral becomes<\/p>\n

\\[
\n\\int
\nu^2\\cdot\\bigl(1+u^2\\bigr)\\,du.
\n\\]<\/p>\n

Expand the expression:<\/p>\n

\\[
\n\\int \\bigl(u^2+u^4\\bigr)\\,du.
\n\\]<\/p>\n

Integrate each term:<\/p>\n

\\[
\n\\int \\bigl(u^2+u^4\\bigr)\\,du
\n=
\n\\frac{u^3}{3}
\n+
\n\\frac{u^5}{5}
\n+
\nC.
\n\\]<\/p>\n

Return to the original variable:<\/p>\n

\\[
\nu=\\tan(x).
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int \\tan^2(x)\\cdot\\sec^4(x)\\,dx
\n=
\n\\frac{\\tan^3(x)}{3}
\n+
\n\\frac{\\tan^5(x)}{5}
\n+
\nC.
\n\\]<\/p>\n

Example 4. Evaluate the integral
\n\\[
\n\\int \\cos(5\\cdot x)\\cdot\\cos(3\\cdot x)\\,dx.
\n\\]<\/h3>\n

The integrand contains a product of cosines with different arguments. Therefore, we use the formula<\/p>\n

\\[
\n\\cos(\\alpha)\\cdot\\cos(\\beta)
\n=
\n\\frac{1}{2}\\cdot
\n\\bigl(
\n\\cos(\\alpha-\\beta)+\\cos(\\alpha+\\beta)
\n\\bigr).
\n\\]<\/p>\n

In our case,<\/p>\n

\\[
\n\\alpha=5\\cdot x,
\n\\qquad
\n\\beta=3\\cdot x.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\n\\cos(5\\cdot x)\\cdot\\cos(3\\cdot x)
\n=
\n\\frac{1}{2}\\cdot
\n\\bigl(
\n\\cos(2\\cdot x)+\\cos(8\\cdot x)
\n\\bigr).
\n\\]<\/p>\n

The original integral becomes<\/p>\n

\\[
\n\\frac{1}{2}\\cdot
\n\\int
\n\\bigl(
\n\\cos(2\\cdot x)+\\cos(8\\cdot x)
\n\\bigr)\\,dx.
\n\\]<\/p>\n

Split it into two integrals:<\/p>\n

\\[
\n\\frac{1}{2}\\cdot
\n\\int\\cos(2\\cdot x)\\,dx
\n+
\n\\frac{1}{2}\\cdot
\n\\int\\cos(8\\cdot x)\\,dx.
\n\\]<\/p>\n

For the first integral, we have<\/p>\n

\\[
\n\\int\\cos(2\\cdot x)\\,dx
\n=
\n\\frac{1}{2}\\cdot\\sin(2\\cdot x).
\n\\]<\/p>\n

For the second integral,<\/p>\n

\\[
\n\\int\\cos(8\\cdot x)\\,dx
\n=
\n\\frac{1}{8}\\cdot\\sin(8\\cdot x).
\n\\]<\/p>\n

Substitute the antiderivatives:<\/p>\n

\\[
\n\\frac{1}{2}\\cdot
\n\\frac{1}{2}\\cdot
\n\\sin(2\\cdot x)
\n+
\n\\frac{1}{2}\\cdot
\n\\frac{1}{8}\\cdot
\n\\sin(8\\cdot x)
\n+
\nC.
\n\\]<\/p>\n

After simplification, we obtain<\/p>\n

\\[
\n\\int \\cos(5\\cdot x)\\cdot\\cos(3\\cdot x)\\,dx
\n=
\n\\frac{1}{4}\\cdot\\sin(2\\cdot x)
\n+
\n\\frac{1}{16}\\cdot\\sin(8\\cdot x)
\n+
\nC.
\n\\]<\/p>\n

Example 5. Evaluate the integral
\n\\[
\n\\int
\n\\frac{dx}
\n{3+\\sin(x)+\\cos(x)}.
\n\\]<\/h3>\n

The denominator contains both sine and cosine. The numerator does not contain a factor that can be directly related to the derivative of either of these functions. Therefore, we use the universal trigonometric substitution<\/p>\n

\\[
\nt=\\tan\\left(\\frac{x}{2}\\right).
\n\\]<\/p>\n

With this substitution,<\/p>\n

\\[
\n\\begin{gathered}
\n\\sin(x)=\\frac{2\\cdot t}{1+t^2},\\\\[4pt]
\n\\cos(x)=\\frac{1-t^2}{1+t^2},\\\\[4pt]
\ndx=\\frac{2}{1+t^2}\\,dt.
\n\\end{gathered}
\n\\]<\/p>\n

Substitute these expressions into the original integral:<\/p>\n

\\[
\n\\int
\n\\frac{
\n\\frac{2}{1+t^2}\\,dt
\n}{
\n3+
\n\\frac{2\\cdot t}{1+t^2}
\n+
\n\\frac{1-t^2}{1+t^2}
\n}.
\n\\]<\/p>\n

Bring the terms in the denominator to a common denominator:<\/p>\n

\\[
\n3+
\n\\frac{2\\cdot t}{1+t^2}
\n+
\n\\frac{1-t^2}{1+t^2}
\n=
\n\\frac{
\n3\\cdot\\bigl(1+t^2\\bigr)
\n+
\n2\\cdot t
\n+
\n1-t^2
\n}{
\n1+t^2
\n}.
\n\\]<\/p>\n

Expand the expression in the numerator:<\/p>\n

\\[
\n3\\cdot\\bigl(1+t^2\\bigr)
\n+
\n2\\cdot t
\n+
\n1-t^2
\n=
\n3+3\\cdot t^2+2\\cdot t+1-t^2.
\n\\]<\/p>\n

After combining like terms, we obtain<\/p>\n

\\[
\n2\\cdot t^2+2\\cdot t+4.
\n\\]<\/p>\n

Factor out the common factor \\( 2 \\):<\/p>\n

\\[
\n2\\cdot t^2+2\\cdot t+4
\n=
\n2\\cdot\\bigl(t^2+t+2\\bigr).
\n\\]<\/p>\n

Therefore, the original integral becomes<\/p>\n

\\[
\n\\int
\n\\frac{
\n\\frac{2}{1+t^2}\\,dt
\n}{
\n\\frac{
\n2\\cdot\\bigl(t^2+t+2\\bigr)
\n}{
\n1+t^2
\n}
\n}.
\n\\]<\/p>\n

Cancel the factors \\( 2 \\) and \\( 1+t^2 \\):<\/p>\n

\\[
\n\\int
\n\\frac{dt}
\n{t^2+t+2}.
\n\\]<\/p>\n

To apply the standard arctangent formula, complete the square in the denominator:<\/p>\n

\\[
\nt^2+t+2
\n=
\n\\left(t+\\frac{1}{2}\\right)^2
\n+
\n\\frac{7}{4}.
\n\\]<\/p>\n

Write \\( 7\/4 \\) as a square:<\/p>\n

\\[
\n\\frac{7}{4}
\n=
\n\\left(\\frac{\\sqrt{7}}{2}\\right)^2.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int
\n\\frac{dt}
\n{
\n\\left(t+\\frac{1}{2}\\right)^2
\n+
\n\\left(\\frac{\\sqrt{7}}{2}\\right)^2
\n}.
\n\\]<\/p>\n

We use the formula<\/p>\n

\\[
\n\\int
\n\\frac{du}
\n{u^2+a^2}
\n=
\n\\frac{1}{a}\\cdot
\n\\arctan\\left(\\frac{u}{a}\\right)
\n+
\nC.
\n\\]<\/p>\n

Let<\/p>\n

\\[
\nu=t+\\frac{1}{2},
\n\\qquad
\ndu=dt,
\n\\qquad
\na=\\frac{\\sqrt{7}}{2}.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\n\\int
\n\\frac{dt}
\n{
\n\\left(t+\\frac{1}{2}\\right)^2
\n+
\n\\left(\\frac{\\sqrt{7}}{2}\\right)^2
\n}
\n=
\n\\frac{2}{\\sqrt{7}}\\cdot
\n\\arctan\\left(
\n\\frac{
\nt+\\frac{1}{2}
\n}{
\n\\frac{\\sqrt{7}}{2}
\n}
\n\\right)
\n+
\nC.
\n\\]<\/p>\n

Simplify the argument of the arctangent:<\/p>\n

\\[
\n\\frac{
\nt+\\frac{1}{2}
\n}{
\n\\frac{\\sqrt{7}}{2}
\n}
\n=
\n\\frac{2\\cdot t+1}{\\sqrt{7}}.
\n\\]<\/p>\n

We obtain<\/p>\n

\\[
\n\\frac{2}{\\sqrt{7}}\\cdot
\n\\arctan\\left(
\n\\frac{2\\cdot t+1}{\\sqrt{7}}
\n\\right)
\n+
\nC.
\n\\]<\/p>\n

Now return to the variable \\( x \\):<\/p>\n

\\[
\nt=\\tan\\left(\\frac{x}{2}\\right).
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int
\n\\frac{dx}
\n{3+\\sin(x)+\\cos(x)}
\n=
\n\\frac{2}{\\sqrt{7}}\\cdot
\n\\arctan\\left(
\n\\frac{
\n2\\cdot\\tan\\left(\\frac{x}{2}\\right)+1
\n}{
\n\\sqrt{7}
\n}
\n\\right)
\n+
\nC.
\n\\]<\/p>\n

The resulting antiderivative should be considered on each interval where the substitution \\( t=\\tan(x\/2) \\) is defined.<\/p>\n

Next Steps in Integral Calculus: Topics to Explore<\/h2>\n

After studying the integration of trigonometric functions, you can move on to other important topics in integral calculus. They will help you work with new types of integrands and make the transition from indefinite integrals to definite integrals.<\/p>\n

    \n
  1. Integrals of Irrational Functions: Common Substitutions<\/a> \u2014 We will look at substitutions that help eliminate radicals and transform irrational expressions into forms that are easier to integrate.<\/li>\n
  2. Definite Integral: Geometric Meaning and Properties<\/a> \u2014 We will explore the geometric meaning of the definite integral, its connection with area, and its main properties.<\/li>\n
  3. Newton\u2013Leibniz Formula: How to Evaluate a Definite Integral<\/a> \u2014 We will learn how to use an antiderivative and the limits of integration to calculate the value of a definite integral.<\/li>\n<\/ol>\n

    Integration of Trigonometric Functions: From Flowchart to Program Code<\/h2>\n

    Do you enjoy programming? Then try turning the mathematical calculations into your own program. Implement the algorithm shown in the flowchart using Pascal<\/em>, Python<\/em><\/a>, JavaScript<\/em>, C++<\/em>, or another programming language of your choice. The program should calculate the value of the integrand<\/p>\n

    \\[
    \nf(x)=\\sin^2(x)\\cdot\\cos^3(x),
    \n\\]<\/p>\n

    approximate the derivative of the antiderivative<\/p>\n

    \\[
    \nF(x)=\\frac{\\sin^3(x)}{3}-\\frac{\\sin^5(x)}{5},
    \n\\]<\/p>\n

    and compare the resulting values. This way, you can apply your knowledge of the integration of trigonometric functions in practice and see how a sequence of mathematical calculations can be transformed into program code.<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

    Integration of trigonometric functions does not always come down to directly using a table of antiderivatives. Often, the integrand first<\/p>\n","protected":false},"author":1,"featured_media":4375,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[562],"tags":[564,539,563,566,580],"class_list":["post-4351","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-integral-calculus","tag-antiderivative","tag-calculus","tag-indefinite-integral","tag-integration-examples","tag-integration-of-trigonometric-functions"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4351","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=4351"}],"version-history":[{"count":21,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4351\/revisions"}],"predecessor-version":[{"id":4374,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4351\/revisions\/4374"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/4375"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=4351"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=4351"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=4351"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}