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  • Prove the following (Exercises 34 to 39) Integral 0 to 1 sin inverse x dx equals pi by 2 minus 1

Prove the following (Exercises 34 to 39)\

    Q39.    \int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1

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Let \ I=\int sin^{-}xdx

Integrating by parts we get

\\ I= sin^{-}x\int 1\cdot dx-\int (\frac{\mathrm{d} (sin^{-}x)}{\mathrm{d} x}\int 1\cdot dx)\\ I=xsin^{-}x+c-\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I=I_{1}-I_{2}

For I2 take 1-x2 = t2, -xdx=tdt

\\I_{2}=\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I_{2}=-\int\frac{1}{t}tdt \\ I_{2}=-t+c\\ I_{2}=-\sqrt{1-x^{2}}+c

[I]_{0}^{1}=[I_{1}-I_{2}]_{0}^{1}\\

\\=[xsin^{-}x-(-\sqrt{1-x^{2}})]_{0}^{1}\\ =[xsin^{-}x+\sqrt{1-x^{2}}]_{0}^{1}\\ =[1\cdot \frac{\pi }{2}+0]-[0+1]\\ =\frac{\pi }{2}-1

Hence Proved

 

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Sayak

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