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  • Find the general solution: dy by dx equals sin inverse x

9. Find the general solution:

        \frac{dy}{dx} = \sin^{-1}x

Answers (1)

best_answer

Given, in the question

\frac{dy}{dx} = \sin^{-1}x

\implies \int dy = \int \sin^{-1}xdx

Now,

\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx

Here, u =\sin^{-1}x and v = 1
 

\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx

 \\ Let\ 1- x^2 = t \\ \implies -2xdx = dt \implies xdx = -dt/2

\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) \\ \implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C \\ \implies y = x\sin^{-1}x + \sqrt{1-x^2} + C

 

Posted by

HARSH KANKARIA

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