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Exercise 9.3 · Q9

Q.Solve the differential equation dydx=sin⁡−1x\dfrac{dy}{dx} = \sin^{-1} x.

Sikkim CbseNCERTSubjective· 3mImportance★★★★★
Appeared in past exams:COMEDK 2024· Set 2024-A· 1mexact
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Integrating the right side by parts gives y=xsin⁡−1x+1−x2+Cy = x\sin^{-1}x + \sqrt{1-x^{2}} + C.

Idea

The equation dydx=sin⁡−1x\dfrac{dy}{dx}=\sin^{-1}x is variables-separated already: yy is just the antiderivative of sin⁡−1x\sin^{-1}x. So we only need to integrate the right-hand side.

Set up

y=∫sin⁡−1x dx.y=\int \sin^{-1}x\,dx.

There is no elementary shortcut for ∫sin⁡−1x dx\int \sin^{-1}x\,dx, so we integrate by parts, treating sin⁡−1x\sin^{-1}x as the part to differentiate.

Work the steps

  1. Choose u=sin⁡−1xu=\sin^{-1}x and dv=dxdv=dx. Then

du=11−x2 dx,v=x.du=\frac{1}{\sqrt{1-x^{2}}}\,dx,\qquad v=x.

  1. Apply ∫u dv=uv−∫v du\int u\,dv = uv-\int v\,du:

∫sin⁡−1x dx=xsin⁡−1x−∫x1−x2 dx.\int \sin^{-1}x\,dx = x\sin^{-1}x-\int \frac{x}{\sqrt{1-x^{2}}}\,dx.

  1. Evaluate the leftover integral with the substitution t=1−x2t=1-x^{2}, so dt=−2x dxdt=-2x\,dx:

∫x1−x2 dx=−12∫dtt=−t=−1−x2.\int \frac{x}{\sqrt{1-x^{2}}}\,dx = -\frac{1}{2}\int \frac{dt}{\sqrt{t}} = -\sqrt{t} = -\sqrt{1-x^{2}}.

  1. Substitute back: …

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