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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integrals\[\int\limits_0^\frac{\pi}{4}\frac{\sin x+\cos x}{9+16\sin 2x}\;dx\]

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  • (i)Let $f(x)=t,\;then \;f'(x)=dt \;Therefore\; \int f(x)dx=\int t.dt$
  • (ii)$\sin 2x=2\sin x\cos x$
  • (iii)$\int \frac{dx}{a^2-x^2}=\frac{1}{2a} log \frac{|a-x|}{|a+x|}+c$
Let I=$\int\limits_0^\frac{\pi}{4}\frac{\sin x+\cos x}{9+16\sin 2x}\;dx$
$=\int\limits_0^\frac{\pi}{4}\frac{\sin x+\cos x}{9+16(2\sin x \cos x)}\;dx$
Let $(\sin x-\cos x)=t$ on differentiating w.r.t x we get , $(\cos x+\sin x)dx=dt$
When we substitute x for t then the limit also changes
When $x=0,t=\sin 0-\cos 0=-1 \qquad(\cos 0=1)$
When $x=\frac{\pi}{4},t=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=0 \qquad(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt 2})$
$(\sin x-\cos x)^2=t^2$
On expanding we get,
$\sin^2x+\cos^2x-2\sin x \cos x=t^2$
$1-2\sin x \cos x =t^2$
$ =>1-\sin 2x=t^2$
Therefore $ \sin 2x=1-t^2$
Now substituting t and dt we get,
$I=\int \limits_{-1}^0 \frac{dt}{9+16(1-t^2)}$
$=\int \limits_{-1}^0 \frac{dt}{25-16t^2}=\int \limits_{-1}^0 \frac{dt}{(5)^2-(4t)^2)}$$
This is of the form $\int \frac{dx}{a^2-x^2}=\frac{1}{2a} log \frac{|a-x|}{|a+x|}+c$
Here $a=5\;and\; x=4t$
Therefore $\int \limits_{-1}^0 \frac{dt}{5^2-(2t)^2}=\frac{-1}{4}\bigg[\frac{1}{2 \times 5} log \frac{|5-4t|}{|5+4t|}\bigg]_{-1}^0 dt$
$=\frac{1}{40}\bigg[ log \frac{|5-4t|}{|5+4t|}\bigg]_{-1}^0 $
On applying limits
$=\frac{1}{40} log \bigg[\bigg|\frac{5}{5}\bigg|- \bigg(-log \bigg|\frac{1}{9}\bigg|\bigg)\bigg]$
$=\frac{1}{40} log(1)-log(1/9)$
But $log(1)=0\qquad and \;-log(1/9)=log 9$
Therefore $I=\frac{1}{40}[log 9]+c$



answered Feb 19, 2013 by meena.p
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