Question

Find the integral of ${\cos }^{4}xdx$.

Answer 1

Step 1: Simplify the given function.

Given function: ${\cos }^{4}x$

We can write the given function as,

${\cos }^{4}x=\left({\cos }^{2}x\right)\left({\cos }^{2}x\right)...\left(1\right)$

We know the double angle $\mathrm{cosine}$ formula as,

$$\begin{gathered} \cos 2\theta =1-2{\cos }^2\theta \\ \Rightarrow 2{\cos }^2\theta =1+\cos 2\theta \\ \Rightarrow {\cos }^2\theta =\frac{1}{2}\left(1+\cos 2\theta \right) \end{gathered}$$

So by applying the above formula in equation $\left(1\right)$ we get,

$$\begin{gathered} {\cos }^{4}x=\frac{1}{2}\left(1+\cos 2x\right)\times \frac{1}{2}\left(1+\cos 2x\right) \\ =\frac{1}{4}\left(1+2\cos 2x+{\cos }^{2}2x\right) \\ =\frac{1}{4}\left(\left(1+2\cos 2x\right)+\left(\frac{1}{2}(1+\cos 4x)\right)\right)\mathbf{}\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\theta }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{2}\mathbf{\theta }\mathbf{)}\mathbf{\right]} \\ =\frac{1}{4}\left(\left(1+2\cos 2x\right)+\frac{1}{2}+\frac{\cos 4x}{2}\right) \\ =\frac{1}{4}+\frac{1}{8}+\frac{2\cos 2x}{4}+\frac{\cos 4x}{8} \\ =\frac{3}{8}+\frac{\cos 2x}{2}+\frac{\cos 4x}{8} \\ \therefore {\cos }^{4}x=\frac{3}{8}+\frac{\cos \left(2x\right)}{2}+\frac{\cos \left(4x\right)}{8} \end{gathered}$$

Step 2: Find the integral.

$$\begin{gathered} \int {\cos }^{4}xdx=\int \left(\frac{3}{8}+\frac{\cos \left(2x\right)}{2}+\frac{\cos \left(4x\right)}{8}\right)dx \\ =\frac{3}{8}\int dx+\frac{1}{2}\int \cos \left(2x\right)dx+\frac{1}{8}\int \cos \left(4x\right)dx \end{gathered}$$

$=\frac{3}{8}\left(x\right)+\frac{1}{2}\left(\frac{\sin \left(2x\right)}{2}\right)+\frac{1}{8}\left(\frac{\sin \left(4x\right)}{4}\right)+C$Whe $C$ is the integration constant. $\mathrm{[}\mathrm{}\mathrm{\because }\mathrm{\int }\cos \left(\mathrm{nx}\right)\mathrm{}\mathrm{dx}\mathrm{=}\frac{\sin \left(\mathrm{nx}\right)}{\mathrm{n}}\mathrm{]}$

$=\frac{3}{8x}+\frac{\sin \left(2x\right)}{4}+\frac{\sin \left(4x\right)}{32}+C$

Hence, $\int {\cos }^{4}xdx=\frac{3}{8x}+\frac{\sin \left(2x\right)}{4}+\frac{\sin \left(4x\right)}{32}+C$Where $C$ is the integration constant.