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Exercise 4.2 · Q21

Q.Evaluate: ∫01/2sin⁡−1x(1−x2)3/2 dx\int_0^{1/\sqrt2} \dfrac{\sin^{-1}x}{(1-x^2)^{3/2}}\,dx

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Since ddx[x1−x2]=1(1−x2)3/2\dfrac{d}{dx}\Big[\dfrac{x}{\sqrt{1-x^2}}\Big]=\dfrac1{(1-x^2)^{3/2}}, take u=sin⁡−1xu=\sin^{-1}x, dv=(1−x2)−3/2dx⇒v=x1−x2dv=(1-x^2)^{-3/2}dx\Rightarrow v=\dfrac{x}{\sqrt{1-x^2}}.

∫sin⁡−1x(1−x2)3/2dx=xsin⁡−1x1−x2−∫x1−x2 dx=xsin⁡−1x1−x2+12ln⁡(1−x2).\int\frac{\sin^{-1}x}{(1-x^2)^{3/2}}dx=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}-\int\frac{x}{1-x^2}\,dx=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac12\ln(1-x^2). …

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