# Solve an example of integral containing trigonometric functions raised to n power?

*print*Print*list*Cite

### 1 Answer

We'll choose the function to be integrated:

f(x) = (sin x)^n

We'll integrate the function:

Int (sin x)^n dx

We'll re-write the given function as a product of (sin x)*(sin x)^n-1.

We'll write (sin x)^(n-1) = [1 - *(cos x)^2]^(n-1)/2.

We'll re-write the integral:

Int (sin x)^n dx = Int [1 - (cos x)^2]^(n-1)/2*(cos x)'dx

We'll give values to n. We'll choose n = 5.

Int (sin x)^5 dx = Int [1 - (cos x)^2]^(5-1)/2*(cos x)'dx

We'll substitute cos x = t

(cos x)'dx = -dt

We'll re-write the integral of the function in t:

Int [1 - (cos x)^2]^(5-1)/2*(cos x)'dx = -Int (1 - t^2)^2dt

We'll expand the square:

-Int (1 - t^2)^2dt = -Int (1 - 2t^2 + t^4) dt

We'll apply the property of additivity of integrals:

Int (1 - t^2)^2dt = -Int dt + 2Int t^2dt - Int t^4dt

Int dt - 2Int t^2dt + Int t^4dt = -t + 2t^3/3 - t^5/5

**Int (sin x)^5 dx = -cos x + 2(cos x)^3/3 - (cos x)^5/5 + C**