Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • $ \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}$
answered Apr 26, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App