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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\int\large\frac{\sin^{-1}\sqrt x-\cos^{-1}\sqrt x}{\sin^{-1}\sqrt x+\cos^{-1}\sqrt x},$$x\;\in\;[0,1]$

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Toolbox:
  • $\sin^{-1}\sqrt x+\cos^{-1}\sqrt x=\large\frac{\pi}{2}$
  • Integration by parts:$\int udv=uv-\int vdu$
Step 1:
Given $I=\large\int\frac{\sin^{-1}\sqrt x-\cos^{-1}\sqrt x}{\sin^{-1}\sqrt x+\cos^{-1}\sqrt x}$$dx$
We know that $\sin^{-1}\sqrt x+\cos^{-1}\sqrt x=\large\frac{\pi}{2}$
$\cos^{-1}\sqrt x=\large\frac{\pi}{2}$$-\sin^{-1}\sqrt x$
$I=\large\frac{\sin^{-1}\sqrt x-(\large\frac{\pi}{2}-\sin^{-1}\sqrt x)}{\Large\frac{\pi}{2}}$$dx$
$\;\;\;=\large\frac{2}{\pi}$$\int\sin^{-1}\sqrt x-\large\frac{\pi}{2})$$dx$
$\;\;\;=\large\frac{4}{\pi}$$\int\sin^{-1}\sqrt xdx-\int dx$
$\;\;\;=\large\frac{4}{\pi}$$\int \sin^{-1}\sqrt xdx-x$
$\Rightarrow I=\large\frac{4}{\pi}$$I_1-x+c$------(1)
Where $I_1=\int \sin^{-1}\sqrt xdx$
Put $\sqrt x=t$
Squaring on both sides we get,
$x=t^2$
Step 2:
Differentiating with respect to $t$
$dx=2tdt$
$I_1=\int \sin^{-1}t 2t dt$
$\;\;\;=2\int \sin^{-1}t .t dt$
$\;\;\;=2[\sin ^{-1}t.\large\frac{t^2}{2}-\int \large\frac{1}{\sqrt{1-t^2}}.\frac{t^2}{2}$$dt]$
Step 3:
Using integration by parts I
$\int uv dx=u\int vdx-\int \large\frac{d}{dx}$$u\int vdx)dx$
$\Rightarrow t^2\sin^{-1}t-\int \large\frac{t^2}{\sqrt{1-t^2}}$$dt$
$\Rightarrow t^2\sin^{-1}t-\int \large\frac{-(1-t^2)+1}{\sqrt{1-t^2}}$$dt$
$\Rightarrow t^2\sin^{-1}t+\int \sqrt{1-t^2}dt-\int \large\frac{1}{\sqrt{1-t^2}}$$dt$
$\Rightarrow t^2\sin^{-1}t+\large\frac{t\sqrt{1-t^2}}{2}+\frac{1}{2}$$\sin^{-1}t-\sin^{-1}t$
$\Rightarrow (t^2-\large\frac{1}{2})$$\sin^{-1}t+\large\frac{1}{2}$$t\sqrt{1-t^2}$
$\Rightarrow \large\frac{1}{2}$$[(2x-1)\sin^{-1}\sqrt x+\sqrt{x-x^2}]$
Step 4:
On putting the value of $I_1$ in equ(1) we get,
$\large\int\frac{\sin^{-1}\sqrt x-\cos^{-1}\sqrt x}{\sin^{-1}\sqrt x+\cos^{-1}\sqrt x}$$dx=\large\frac{2}{\pi}$$[(2x-1)\sin^{-1}\sqrt x+\sqrt{x-x^2}]+c$
answered Sep 10, 2013 by sreemathi.v
 
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