Aspire Faculty ID #3637 · Topic: NIMCET 2012 · Just now
NIMCET 2012

The value of $\displaystyle \int_{0}^{\sin^2 x} \sin^{-1}\sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1}\sqrt{t} dt$ is:

Solution

Use the identity:
 $\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}$ 
So: $\cos^{-1}\sqrt{t}=\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}$ 
Now substitute: $I=\displaystyle\int_0^{\sin^2 x}\sin^{-1}\sqrt{t}  dt + \int_0^{\cos^2 x} \left(\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}\right) dt$ 
$I=\dfrac{\pi}{2}\cos^2 x + \int_0^{\sin^2 x}\sin^{-1}\sqrt{t}dt - \int_0^{\cos^2 x}\sin^{-1}\sqrt{t} dt$ 
Combine integrals: $I=\dfrac{\pi}{2}\cos^2 x + \int_{\cos^2 x}^{\sin^2 x}\sin^{-1}\sqrt{t} dt$ 
But: $\sin^2 x + \cos^2 x = 1$ 
Limits become from $1$ to $0$: 
$I=\dfrac{\pi}{2}(1 - \sin^2 x) + \int_{1}^{0}\sin^{-1}\sqrt{t} dt$ 

$I=\dfrac{\pi}{2} - \int_0^{1}\sin^{-1}\sqrt{t} dt$ 

Let $u=\sqrt{t}$, 
$dt=2udu$: 

$\displaystyle \int_0^{1}\sin^{-1}\sqrt{t} dt = 2\int_0^{1} u\sin^{-1}udu$ 

Standard result: $\displaystyle 2\int_0^{1} u\sin^{-1}u du = \dfrac{\pi}{4}$ 

Thus: $I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$
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