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Evaluate: $\int\limits_0^ a \frac{dx}{x+\sqrt {a^2-x^2}}$

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  • $ \int \limits_a^b f(x)dx=F(b)-F(a)$
  • $(ii)\; \int \limits_0^a f(x) dx= \int \limits_0^a f(a-x)dx$
given $ I=\int \limits_0^a\large \frac {dx}{x+\sqrt {a^2-x^2}}$
put $x=a \sin \theta$ on differentiating w.r.t x,
$dx= a\cos \theta d\theta$
The limits also change as we substitute $x=a \sin \theta,$
When $x=0, a\sin \theta=0 =>\theta=0$
When $x=a,a \sin \theta=a=>\sin \theta=1=>\theta=\pi/2$
On substituting $\sin \theta\; and\; d \theta$,
Therefore $I=\int \limits_0^{\pi/2} \large\frac{a \cos \theta \;d \theta}{a \sin \theta+\sqrt {a^2-a^2\sin ^2 \theta}}$
$=\int \limits_0^{\pi/2} \large\frac{a \cos \theta \;d \theta}{a \sin \theta+a\sqrt {1-\sin ^2 \theta}}\qquad$ But $ \sqrt {1-\sin ^2 \theta}=\cos \theta$
Therefore $I=\int \limits_0^{\pi/2} \large\frac{a \cos \theta \;d \theta}{a \sin \theta+a \cos \theta}$------(1)
By applying the property $ \int \limits_0^a f(x) dx= \int \limits_0^a f(a-x)dx$
$I=\int \limits_0^{\pi/2} \large\frac{a \cos (\pi/2-\theta) \;d \theta}{a \sin (\pi/2-\theta)+a \cos (\pi/2-\theta)}$
But $\sin(\pi/2-\theta)=\cos \theta\;and\; \cos (\pi/2-\theta)=\sin \theta$
Therefore $I=\int \limits_0^{\pi/2} \large\frac{a \sin \theta \;d \theta}{a \cos \theta+a \sin \theta}$------(2)
Add equ (1) and equ (2)
$2I=\int \limits_0^{\pi/2}\bigg( \large\frac{a \sin \theta}{a \cos \theta+a \sin \theta}+\large\frac{a \cos \theta }{a \cos \theta+a \sin \theta}\bigg)$$d \theta$
$=\int \limits_0^{\pi/2} \large\frac{a \sin \theta+a \cos \theta}{a \cos \theta+a \sin \theta}$$d\theta$
$2I=\int \limits_0^{\pi/2} d\theta$
On integrating we get
$2I= \bigg[0\bigg]_0^{\pi/2}$
On applying the limits,
Therefore $ I=\large\frac{\pi}{4}$
answered May 6, 2013 by meena.p