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Primitive of sin^-1x

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Primitive of sin^-1x

Posted by Akshat Mehta 6 years, 9 months ago

CBSE > Class 12 > Mathematics

1 answers

Payal Singh 6 years, 9 months ago

{tex}\int sin^{-1}x \ dx{/tex}

= {tex}\int sin^{-1}x.1 \ dx{/tex}

Now Integrate by parts, Using ILATE first Function is sin^-1 x and 2nd Function is 1.

= {tex}sin^{-1}x\int 1\ dx- \int {1\over \sqrt {1-x^2}}xdx{/tex}

{tex}= sin^{-1}x.x- I_1{/tex} .....(1)

Where {tex}I_1 = \int {1\over \sqrt {1-x^2}}xdx {/tex}

Put {tex}{1-x^2} = t{/tex}

-2xdx = dt

{tex}=> I_1 = -{1\over 2}\int {1\over \sqrt t}dt{/tex}

{tex}=> I_1 = - {\sqrt t}{/tex}

{tex}=> I_1 = -{\sqrt {1-x^2}}{/tex}

Put in (1), we get

= {tex}sin^{-1}x.x+\sqrt{1-x^2} + c{/tex}

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