Trigonometric Integrals

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Trigonometric Integrals

In this topic, we will study how to integrate certain combinations involving products and powers of trigonometric functions.

We consider 8 cases.

1. Integrals of the form cosaxcosbxdx, sinaxcosbxdx, sinaxsinbxdx

To evaluate integrals of products of sine and cosine with different arguments, we apply the identities

cosaxcosbx=12[cos(ax+bx)+cos(axbx)];
sinaxcosbx=12[sin(ax+bx)+sin(axbx)];
sinaxsinbx=12[cos(ax+bx)cos(axbx)].

2. Integrals of the form sinmxcosnxdx

We assume here that the powers m and n are non-negative integers.

To find an integral of this form, use the following substitutions:

  1. If m (the power of sine) is odd, we use the usubstitution
    u=cosx,du=sinxdx

    and the identity

    sin2x+cos2x=1

    to express the remaining even power of sine in uterms.

  2. If n (the power of cosine) is odd, we use the usubstitution
    u=sinx,du=cosxdx

    and the identity

    sin2x+cos2x=1

    to express the remaining even power of cosine in uterms.

  3. If both powers m and n are even, we reduce the powers using the half-angle formulas
    sin2x=1cos2x2,cos2x=1+cos2x2.

The integrals of type sinnxdx and cosnxdx can be evaluated by reduction formulas

sinnxdx=sinn1xcosxn+n1nsinn2xdx,
cosnxdx=cosn1xsinxn+n1ncosn2xdx.

3. Integrals of the form tannxdx

The power of the integrand can be reduced using the trigonometric identity

1+tan2x=sec2x

and the reduction formula

tannxdx=tann2xtan2xdx=tann2x(sec2x1)dx=tann1xn1tann2xdx.

4. Integrals of the form cotnxdx

The power of the integrand can be reduced using the trigonometric identity

1+cot2x=csc2x

and the reduction formula

cotnxdx=cotn2xcot2xdx=cotn2x(csc2x1)dx=cotn1xn1cotn2xdx.

5. Integrals of the form secnxdx

This type of integrals can be simplified with help of the reduction formula:

secnxdx=secn2xtanxn1+n2n1secn2xdx.

6. Integrals of the form cscnxdx

Similarly to the previous examples, this type of integrals can be simplified by the formula

cscnxdx=cscn2xcotxn1+n2n1cscn2xdx.

7. Integrals of the form tanmxsecnxdx

  1. If the power of the secant n is even, then using the identity
    1+tan2x=sec2x

    the secant function is expressed as the tangent function. The factor sec2x is separated and used for transformation of the differential. As a result, the entire integral (including differential) is expressed in terms of the function tanx.

  2. If both the powers n and m are odd, then the factor secxtanx, which is necessary to transform the differential, is separated. Then the entire integral is expressed in terms of secx.
  3. If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is expressed in terms of the secant using the identity
    1+tan2x=sec2x.

    After this substitution, you can calculate the integrals of the secant.

8. Integrals of the form cotmxcscnxdx

  1. If the power of the cosecant n is even, then using the identity
    1+cot2x=csc2x

    the cosecant function is expressed as the cotangent function. The factor csc2x is separated and used for transformation of the differential. As a result, the integrand and differential are expressed in terms of cotx.

  2. If both the powers n and m are odd, then the factor cotxcscx, which is necessary to transform the differential, is separated. Then the integral is expressed in terms of cscx.
  3. If the power of the cosecant n is odd, and the power of the cotangent m is even, then the cotangent is expressed in terms of the cosecant using the identity
    1+cot2x=csc2x.

    After this substitution, you can find the integrals of the cosecant.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Calculate the integral sin3xdx.

Example 2

Evaluate the integral cos5xdx.

Example 3

Find the integral sin6xdx.

Example 4

Find the integral sin2xcos3xdx.

Example 5

Calculate the integral sin2xcos4xdx.

Example 6

Evaluate the integral sin3xcos4xdx.

Example 7

Evaluate the integral sin3xcos5xdx.

Example 8

Evaluate the integral sin3xcos3xdx.

Example 1.

Calculate the integral sin3xdx.

Solution.

Let u=cosx, du=sinxdx. Then

sin3xdx=sin2xsinxdx=(1cos2x)sinxdx=(1u2)du=(u21)du=u33u+C=cos3x3cosx+C.

Example 2.

Evaluate the integral cos5xdx.

Solution.

Making the substitution u=sinx, du=cosxdx and using the identity

cos2x=1sin2x,

we obtain:

cos5xdx=(cos2x)2cosxdx=(1sin2x)2cosxdx=(1u2)2du=(12u2+u4)du=u2u33+u55+C=sinx2sin3x3+sin5x5+C.

Example 3.

Find the integral sin6xdx.

Solution.

Using the identities

sin2x=1cos2x2andcos2x=1+cos2x2,

we can write:

I=sin6xdx=(sin2x)3dx=18(1cos2x)3dx=18(13cos2x+3cos22xcos32x)dx=x838sin2x2+38cos22xdx38cos32xdx.

Calculate the integrals in the latter expression.

cos22xdx=1+cos4x2dx=12(1+cos4x)dx=12(x+sin4x4)=x2+sin4x8.

To find the integral cos32xdx, we make the substitution u=sin2x, du= 2cos2xdx. Then

u=sin2x,du=2cos2xdx.

Then

cos32xdx=122cos22xcos2xdx=122(1sin22x)cos2xdx=12(1u2)du=u2u36=sin2x2sin32x6.

Hence, the initial integral is

I=x83sin2x16+38(x2+sin4x8)18(sin2x2sin32x6)+C=5x16sin2x4+3sin4x64+sin32x48+C.

Example 4.

Find the integral sin2xcos3xdx.

Solution.

The power of cosine is odd, so we make the substitution

u=sinx,du=cosxdx.

We rewrite the integral in terms of sinx to obtain:

sin2xcos3xdx=sin2xcos2xcosxdx=sin2x(1sin2x)cosxdx=u2(1u2)du=(u2u4)du=u33u55+C=sin3x3sin5x5+C.

Example 5.

Calculate the integral sin2xcos4xdx.

Solution.

We can write:

I=sin2xcos4xdx=(sinxcosx)2cos2xdx.

We convert the integrand using the identities

sinxcosx=sin2x2,cos2x=1+cos2x2,sin2x=1cos2x2.

This yields

I=(sin2x2)21+cos2x2dx=18sin22x(1+cos2x)dx=18sin22xdx+18sin22xcos2xdx=181cos4x2dx+1162sin22xcos2xdx=116(1cos4x)dx+116sin22xd(sin2x)=116(xsin4x4)+116sin32x3+C=x16sin4x64+sin32x48+C.

Example 6.

Evaluate the integral sin3xcos4xdx.

Solution.

As the power of sine is odd, we use the substitution

u=cosx,du=sinxdx.

The integral is written as

I=sin3xcos4xdx=sin2xcos4xsinxdx.

By the Pythagorean identity,

sin2x=1cos2x.

Hence

I=sin2xcos4xsinxdx=(1cos2x)cos4xsinxdx=(1u2)u4du=(u6u4)du=u77u55+C=cos7x7cos5x5+C.

Example 7.

Evaluate the integral sin3xcos5xdx.

Solution.

We see that both powers are odd, so we can substitute either u=sinx or u=cosx. Choosing the least exponent, we have

u=cosx,du=sinxdx.

The integral takes the form

I=sin3xcos5xdx=sin2xcos5xsinxdx.

Using the Pythagorean identity

sin2x=1cos2x,

we can write

I=sin2xcos5xsinxdx=(1cos2x)cos5xsinxdx=(1u2)u5du=(u7u5)du=u88u66+C=cos8x8cos6x6+C.

Example 8.

Evaluate the integral sin3xcos3xdx.

Solution.

The powers of both sine and cosine are odd. Hence we can use the substitution u=sinx or u=cosx. Let's apply the substitution u=sinx. Then du=cosxdx, and the integral becomes

I=sin3xcos3xdx=sin3xcos2xcosxdx.

By the Pythagorean identity,

sin2x=1cos2x,

so we obtain

I=sin3xcos2xcosxdx=sin3x(1sin2x)cosxdx=u3(1u2)du=(u3u5)du=u44u66+C=sin4x4sin6x6+C.