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Question 103 of 129

Q.Evaluate: ∫esin⁡−1x1−x2 dx\displaystyle\int \dfrac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}\,dx.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2018Subjective· 2mImportance★★★★★
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With t=sin⁡−1xt=\sin^{-1}x, dt=dx1−x2dt=\dfrac{dx}{\sqrt{1-x^2}}, so the integral reduces to ∫et dt=et+c=esin⁡−1x+c\int e^t\,dt = e^t+c = e^{\sin^{-1}x}+c.

Let t=sin⁡−1xt = \sin^{-1}x. Then dt=11−x2 dxdt = \dfrac{1}{\sqrt{1-x^2}}\,dx.

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