Ask Questions, Get Answers

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

Evaluate:$\int \limits _0^{\pi/2} \frac{1}{2\cos x+4 \sin x}dx$

Can you answer this question?

1 Answer

0 votes
  • (i)$\int \limits _a^b f(x)dx=F(b)-F(a)$
  • (ii)If we substitute a function f(x) as t, then $f'(x)dx=dt$ hence the function $\int f(x)dx=\int t.dt$
  • (iii)$ \large\frac{dx}{x^2-a^2}=\frac{1}{2a} log \frac{|x-a|}{|x+a|}$
given $I=\large\int \limits _0^{\pi/2} \frac{dx}{2\cos x+4 \sin x}$
Put $\tan x/2=t$ on differentiating w.r.t x we get
$\frac{1}{2} \sec^2 x/2 dx=dt \qquad=> \sec ^2 x/2 dx=dx/2$
Therefore $dx=\large\frac{2dt}{\sec^2 x/2}$ But $ \sec^2 x/2=1+\tan^2 x/2$
Therefore $dx=\large\frac{2dt}{1+t^2}\qquad \cos x=\frac{1-\tan^2 x/2}{1+\tan^2 x/2}=\frac{1-t^2}{1+t^2}$
$\sin x=\large\frac{2\tan^2 x/2}{1+\tan^2 x/2}=\frac{2t}{1+t^2}$
This limits also change
when x=0,t=0; when $x=\pi/2,t=\tan \pi/4=1$
Hence now substituting for $\cos x,\sin x$ and dx we get,
$I=\Large\int \limits_0^1 \frac{\int \limits_0^2 \frac{2dt}{1+t^2}}{2 \bigg(\frac{1-t^2}{1+t^2}\bigg)+4 \bigg(\frac{2t}{1+t^2}\bigg)}$
$=\large \int \limits_0^1 \frac{2dt}{2(1-t^2)+4(2t)}=\int \limits_0^1 \frac{2 dt}{2-2t^2+8t}$
Therefore $I=2\int \limits_0^1 \large\frac{dt}{-2t^2+8t+2}=\frac{2}{-2} \int \frac{dt}{t^2-4t-1}$
$=-\int \limits_0^1 \large\frac{dt}{(t-2)^2-4-1}=-\int \limits_0^1 \frac{dt}{(t-2)^2-(\sqrt 5)^2}$
$I=\int \limits_0^1 \large\frac{dt}{(\sqrt 5)^2-(t-2)^2}$
This is of the form $\int \frac{dx}{a^2-x^2}=\frac{1}{2a} log |\frac{a+x}{a-x}|$ here $a=\sqrt 5\; x=(t-2)$
Hence on integrating we get,
$I=\large\frac{1}{2 \sqrt 5} log \bigg[\frac{\sqrt 5+t-2}{\sqrt 5-t+2} \bigg]_0^1$
Applying the limits we get,
$I=\frac{1}{2 \sqrt 5} \bigg\{log \bigg[\frac{\sqrt 5+1-2}{\sqrt 5-1+2} \bigg]-log \bigg[\frac{\sqrt 5+0-2}{\sqrt 5-0+2} \bigg]\bigg\}$
$I=\frac{1}{2 \sqrt 5} \bigg\{\bigg[log \frac{\sqrt 5-1}{\sqrt 5+1} \bigg]-log \bigg[\frac{\sqrt 5-2}{\sqrt 5+2} \bigg]\bigg\}$
But $ log a-log b =log |a/b|$
Therefore $I=\frac{1}{2 \sqrt 5} \bigg\{\bigg[log \bigg(\frac{\sqrt 5-1}{\sqrt 5+1} \times \frac{\sqrt 5+2}{\sqrt 5-2}\bigg) \bigg]\bigg\}$
$=\frac{1}{2 \sqrt 5} \bigg\{\bigg[log \bigg(\frac{3+\sqrt 5}{3-\sqrt 5}\bigg) \bigg]\bigg\}$
Multiplying and dividing by its conjugate
$I=\frac{1}{2 \sqrt 5} \bigg[log \bigg(\frac{3+\sqrt 5}{3-\sqrt 5} \times \frac{3+\sqrt 5}{3+\sqrt 5}\bigg) \bigg]$
$=\frac{1}{2 \sqrt 5} \bigg[log \frac{(3+\sqrt 5)^2}{3^2-(\sqrt 5)^2}\bigg]=\frac{1}{2 \sqrt 5} log \bigg[\bigg(\frac{3+\sqrt 5}{4}\bigg)^2\bigg]$
But $log a^2=2 log a$
Hence $I=\frac{2}{2 \sqrt 5} \bigg[log \frac{3+\sqrt 5}{2}\bigg]$
Therefore $I=\frac{1}{\sqrt 5} log \bigg(\frac{3+\sqrt 5}{2}\bigg)$
answered Apr 1, 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