Comment
Share
Q)

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

Comment
A)
Toolbox:
• $\int \limits_a^b f(x)dx =F(b)-F(a)$
• $\int \limits_{-a}^a f(x)dx=2 \int \limits_0^a f(x)dx$ if $f(x)$ is an even function
• A function is an even function if $f(-x)=f(x)$
Step 1:
Given $I= \int \limits_{-\pi/2}^{\pi/2} \cos ^4 x dx$
if x is replaced by $-x$ then $\int \limits_{-\pi/2}^{\pi/2} \cos ^4 (-x) dx$$=\int \limits_{-\pi/2}^{\pi/2} \cos ^4 x dx. Hence it is an even function Hence \int \limits_{-\pi/2}^{\pi/2} \cos ^4 x dx=2 \int \limits_0^{\pi/2} \cos ^4 x dx \int \limits_0^{\pi/2} \cos ^4 x dx=\int \limits_0^{\pi/2} (\cos ^2 x)^2 dx But we know \cos ^2 x=\bigg(\large\frac{1+\cos 2x}{2}\bigg)^2 On expanding this we get, \large\frac{1}{2}$$ (1+ \cos ^2 2x+2 \cos 2x)$
Therefore $I= \large\frac{1}{4} $$\int \limits_0^{\pi/2} (1+\cos ^2 2x+2 \cos 2x)dx But again \cos ^2 2x=\large\frac{1+\cos 4x}{2} Therefore I= \large\frac{1}{4}$$\int \limits_0^{\pi/2} dx + \large\frac{1}{8} $$\int \limits_0^{\pi/2} 1+\cos 4x dx+2 . \large\frac{1}{4}$$\int \cos 2x dx$
Step 2:
On integrating seperately
$=\large\frac{1}{4}$$\bigg[x\bigg]_0^{\pi/2}+\large\frac{1}{8}$$\bigg[x+ \large\frac{\sin 4x}{4} $$\bigg]_0^{\pi/2}+\large\frac{1}{2} \bigg[\frac{\sin 2x}{2} \bigg]_0^{\pi/2} Step 3: On applying limits, \large\frac{1}{4}$$\bigg[x\bigg]_0^{\pi/2}+\large\frac{1}{8}$$\bigg[x+\large\frac{1}{4}$$\sin 4x\bigg]_0^{\pi/2}+\large\frac{1}{2} \bigg[\frac{\sin 2x}{2} \bigg]_0^{\pi/2}$
$=\large\frac{1}{4} \bigg[\frac{\pi}{2}-0\bigg]+\frac{1}{8} $$\bigg[ (\pi/2-0)+\large\frac{1}{4}$$(\sin 4.\pi/2-\sin 0)\bigg]+\large\frac{1}{2}\bigg[\frac{1}{2}$$(\sin 2\pi/2-\sin 0)\bigg] =\large\frac{1}{4}$$ (\pi/2)+\large\frac{1}{8} $$(\pi/2)+ \large\frac{1}{32}$$ (\sin 2 \pi -\sin 0)+\large\frac{1}{4}$$(\sin \pi-\sin 0)$
But we know $\sin 2 \pi =0 \;and\; \sin 0=0$
Therefore $I= \large\frac{\pi}{8}+\frac{\pi}{16}=\frac{2 \pi+\pi}{16}=\frac{3\pi}{16}$