Evaluate I=\int\cos^4(x)dx?

Symbolab says to use integration by parts. However, this question originates in a "Trigonometric Integrals" packet... so how would I go about solving this?

Apr 9, 2018

$I = \frac{1}{32} \left(12 x + \sin \left(4 x\right) + 8 \sin \left(2 x\right)\right) + C$

Explanation:

We want to solve

$I = \int {\cos}^{4} \left(x\right) \mathrm{dx}$

I always looking to reduce the powers of the integrand,
for integrals of this type

You could use the identity

color(blue)(cos^2(x)=1/2(1+cos(2x))

Thus

$I = \int {\left({\cos}^{2} \left(x\right)\right)}^{2} \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \int {\left(1 + \cos \left(2 x\right)\right)}^{2} \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \int 1 + {\cos}^{2} \left(2 x\right) + 2 \cos \left(2 x\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \int 1 + \frac{1}{2} \left(1 + \cos \left(4 x\right)\right) + 2 \cos \left(2 x\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \int \frac{3}{2} + \frac{1}{2} \cos \left(4 x\right) + 2 \cos \left(2 x\right) \mathrm{dx}$

Which is much nicer to integrate

$I = \frac{1}{4} \left(\frac{3}{2} x + \frac{1}{8} \sin \left(4 x\right) + \sin \left(2 x\right)\right) + C$

$\textcolor{w h i t e}{I} = \frac{1}{32} \left(12 x + \sin \left(4 x\right) + 8 \sin \left(2 x\right)\right) + C$