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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following: $\int\limits_0^{\Large \frac{1}{2}}\frac{dx}{(1+x^2)\sqrt {1-x^2}}$$(Hint:let\;x=\sin\theta)$

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Toolbox:
  • $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} (x/a)+c$
  • $(1+\tan ^2 x)=\sec^2 x$
  • If $f(x)$ is substituted by $f(t),$ then $f'(x)dx=f'(t)dt,$ hence $\int f(x)dx=\int f(t)dt$
Let $I=\int\limits_0^{\Large \frac{1}{2}}\large\frac{dx}{(1+x^2)\sqrt {1-x^2}}$
Put $x=\sin \theta$, on differentiating we get $dx=\cos \theta d\theta$
When we substitute for x , the limits also changes, $x \sin \theta =>\theta=\sin ^{-1}(x)$
When $x=\large\frac{1}{2}; \;$$\theta=\sin ^{-1}(1/2)=\pi/6$
When $x=0=> \theta=\sin ^{-1}(0)=0$
On substituting this we get
Hence $I=\large\int \limits_0^{\pi/6}\frac{\cos \theta d\theta}{1+\sin ^2 \theta(\sqrt {1-\sin ^2 \theta})}$
But $1-\sin ^2 \theta=\cos ^2 \theta$
Hence $I=\large\int \limits_0^{\pi/6}\frac{\cos \theta d\theta}{(1+\sin ^2 \theta).\cos \theta}$
$=\large\int \limits_0^{\pi/6}\frac{1}{1+\sin ^2 \theta}d \theta$
Divide the numerator and denominator by $\cos ^2 \theta$
$I=\large\int \limits_0^{\pi/6}\frac{\frac{1}{\cos ^2 \theta }}{\Large\frac{1}{\cos^2 \theta}+\frac{\sin ^2 \theta}{\cos ^2 \theta}}$$d \theta$
Hence $I=\large\int \limits_0^{\pi/6} \frac{\sec^2 \theta}{\sec^2 \theta+\tan ^2 \theta} d\theta$
Put $(1+\tan ^2 \theta)=\sec ^2 \theta$ in the denominator;
$I=\int \limits_0^{\pi/6} \large \frac{\sec ^2 \theta d \theta}{(1+\tan ^2 \theta)+\tan ^2\theta}=\int \limits_0^{\pi/6} \frac{\sec^2 \theta d \theta}{1+2 \tan ^2 \theta}$
Put $\tan \theta=t$
On differentiating with respect to t we get, $\sec ^2 \theta d \theta=dt$
Now the limit again changes when we substitute $\theta$ for t
When $\theta=\pi/6\qquad \tan \theta=\tan \pi/6=\large\frac{1}{\sqrt 3} => t=\frac{1}{\sqrt 3}$
When $\theta=0\qquad \tan \theta=\tan 0 => t=0$
Therefore $I=\int \limits_0^{\frac{1}{\sqrt 3}} \large \frac{dt}{1+2t^2}=\int \limits_0^{\frac{1}{\sqrt 3}} \frac{dt}{2(\frac{1}{2}+t^2)}$
This is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} (x/a)$
On integrating we get, where $x=t$ and $ a=\frac{1}{\sqrt 2}$
$I=\large\frac{1}{2}\int \limits_0^{\frac{1}{\sqrt 3}} \large \frac{dt}{(\frac{1}{\sqrt 2})^2+t^2}$$=\large\frac{1}{2} \bigg[\frac{1}{\frac{1}{\sqrt 2}} $$\tan ^{-1} \bigg(\frac{t}{\large\frac{1}{\sqrt 2}}\bigg)\bigg]_0^{\frac{1}{\sqrt 3}}$
$=\large\frac{\sqrt 2}{2} $$\bigg[\tan ^{-1}( \sqrt 2 t)\bigg]_0^{\frac{1}{\sqrt 3}}$
Now applying limits we get
$\large\frac{\sqrt 2}{2} $$\bigg[\tan ^{-1}\bigg( \frac{\sqrt 2}{3}\bigg)-0\bigg]=\large\frac{\sqrt 2}{2} $$\tan ^{-1} \large\frac{\sqrt 2}{3}$
answered Apr 20, 2013 by meena.p
 
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