# Evaluate: $\int\limits _{-\pi/2} ^{\pi/2} \cos ^4x dx$

$\begin{array}{1 1} \frac{3 \pi}{8} \\ \frac{2 \pi}{7} \\ \frac{\pi}{8} \\ \frac{3 \pi}{4}\end{array}$

Toolbox:
• $(i)\; \int \limits_a^b f(x)dx =F(b)-F(a)$
• $(ii)\; \int \limits_{-a}^a f(x)dx=2 \int \limits_0^a f(x)dx \;if \;f(x)$ is an even function
• $f(x)$ is an even function if $f(-x)=f(x)$
Step 1:
$I=\int \limits_{-\pi/2}^{\pi/2} \cos ^4 x\;dx$
Let $f(x)=\cos ^4 x$
$f(-x)= \cos ^4(-x) =\cos ^4 x$
Hence it is an even function
Therefore $I= 2\int \limits_0^{\pi/2} \cos ^4 x\;dx$
$=2 \int \limits_0^{\pi/2} (\cos ^2 x)^2 dx$
But we know $\cos ^2 x =\large\frac{1+\cos 2x}{2}$
Therefore $I=2 \int \limits_0^{\pi/2} \bigg(\large\frac{1+\cos 2x}{2}\bigg)^2 dx$
$=\large\frac{2}{4} $$\int \limits _0^{\pi/2} (1+\cos 2x)^2 dx On expanding, I= \large\frac{1}{2}$$\int \limits_0^{\pi/2}(1+ 2 \cos 2 x+\cos ^2 2x)dx$
Again $\cos ^2 2x$ can be written as
$\large\frac{1+\cos 4x}{2}$
Step 2:
Therefore $I=\large\frac{1}{2} $$\int \limits_0^{\pi/2} (1+ 2 \cos 2x+\bigg(\large\frac{1+\cos 4x}{2}\bigg) dx On seperating the terms we get I= \large\frac{1}{2}$$\int \limits_0^{\pi/2} (1+2 \cos 2x+\large\frac{1}{2}+\frac{\cos 4x}{2} )dx$
$=\large\frac{1}{2}$$\bigg[\int \limits_0^{\pi/2}(1+1/2)dx + 2 \int \limits_0^{\pi/2} \cos 2x dx+ \large\frac{1}{2}$$\int \limits_0^{\pi/2} cos 4x dx\bigg]$
$=\large\frac{1}{2}$$\bigg[\int \limits_0^{\pi/2}3/2dx + 2 \int \limits_0^{\pi/2} \cos 2x dx+ \frac{1}{2} \int \limits_0^{\pi/2} cos 4x dx\bigg] Step 3: On integrating we get =\large\frac{1}{2}\bigg\{\bigg[\large\frac{3}{2}$$ x\bigg]_0^{\pi/2}+2 \bigg[ \large\frac{\sin 2x}{2} $$\bigg]_0^{\pi/2}+\large\frac{1}{2}\bigg[ \frac{\sin 4x}{4}$$\bigg]_0^{\pi/2} \bigg\}$
On applying limits
$I=\large\frac{3}{4}\bigg[\frac{\pi}{2}$$-0 \bigg]+\large\frac{1}{2}$$ \bigg[ \sin 2. \pi/2-\sin 0\bigg]+\large\frac{1}{8} $$\bigg[\sin 4. \large\frac{\pi}{2}$$-\sin 0\bigg]$
$=\large\frac{3\pi}{8}+\frac{1}{2}$$[\sin \pi -\sin 0]+\large\frac{1}{8}$$[\sin 2 \pi -\sin 0]$
But we know $\sin \pi=\sin 0=\sin 2 \pi =0$
Therefore $I=\large\frac{3 \pi}{8}$
answered Apr 29, 2013 by