# Thread: Need help evaluating this integral.

1. ## Need help evaluating this integral.

$\int cos^7(5x)\dif dx$

This problem coincides with my Calculus II course chapter on trigonometric integrals. I am unsure if I should be using the reduction formula for cosine or perhaps rewriting cosine as an even power and using the identity $cos^2(x)+sin^2(x)=1$ and then u substitution.

2. ## Re: Need help evaluating this integral.

Hey jamesrb.

You may want to look at a reduction formula. You could expand everything out to get things in terms of sines and cosines (even squared ones) but there is a relationship for integrating powers of trig functions.

3. ## Re: Need help evaluating this integral.

Originally Posted by jamesrb
$\int cos^7(5x)\dif dx$

This problem coincides with my Calculus II course chapter on trigonometric integrals. I am unsure if I should be using the reduction formula for cosine or perhaps rewriting cosine as an even power and using the identity $cos^2(x)+sin^2(x)=1$ and then u substitution.
\displaystyle \begin{align*} \int{\cos^7{(5x)}\,dx} &= \frac{1}{5} \int{ 5\cos^7{(5x)}\,dx } \\ &= \frac{1}{5} \int{ 5\cos{(5x)}\cos^6{(5x)}\,dx } \\ &= \frac{1}{5}\int{ 5\cos{(5x)}\left[ \cos^2{(5x)} \right] ^3 \, dx } \\ &= \frac{1}{5}\int{ 5\cos{(5x)} \left[ 1 - \sin^2{(5x)} \right] ^3 \, dx } \end{align*}

Now let \displaystyle \begin{align*} u = \sin{(5x)} \implies du = 5\cos{(5x)}\,dx \end{align*} and the integral becomes

\displaystyle \begin{align*} \frac{1}{5}\int{ 5\cos{(5x)} \left[ 1 - \sin^2{(5x)} \right] ^3 \, dx } &= \frac{1}{5} \int{ \left( 1 - u^2 \right)^3 \, du } \\ &= \frac{1}{5} \int{ 1 - 3u^2 + 3u^4 - u^6 \, du } \\ &= \frac{1}{5} \left( u - u^3 + \frac{3u^5}{5} - \frac{u^7}{7} \right) + C \\ &= \frac{1}{5} \left[ \sin{(5x)} - \sin^3{(5x)}+ \frac{3 \sin^5{(5x)}}{5} - \frac{\sin^7{(5x)}}{7} \right] + C \end{align*}

4. ## Re: Need help evaluating this integral.

Generally speaking, any time you have an odd power of cos(x) or sin(x), you can factor out one, leaving an even power. Then use " $cos^2(x)= 1- sin^2(x)$" or " $sin^2(x)= 1- cos^2(x)$ to change to the "other" function, keeping that first term you factored out for the derivative:
$\int cos^{2n+1}(x)dx \int cos^{2n}(x)(cos(x)dx)$ $= \int (cos^2(x))^n(cos(x)dx)= \int (1- sin^2(x))^n (cos(x)dx)$
Now, let u= sin(x) so that du= cos(x)dx and the integral becomes $\int (1- u^2)^n du$.

If you have only even powers of sine and/or cosine, then you will need $sin^2(x)= (1/2)(1- cos(x/2)$ and $cos^2(x)= (1/2)(1+ cos(x/2)$.