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  • Confused! kindly explain, - Integral Calculus - JEE Main-14

If I_{1} = \int \frac{dx}{\sqrt{a^2 - x^2}} and I_{2} = \int \cos^{-1} \frac{x}{a} then I_{1}+ I_{2} = ?

  • Option 1)

    \sin^{-1} \frac{x}{a} +c

  • Option 2)

    \cos^{-1}\frac{x}{a} + c

  • Option 3)

    C

  • Option 4)

    None of these

 

Answers (1)

best_answer

As we have learnt,

 

Integrals of inverse circular functions -

\frac{\mathrm{d} }{\mathrm{d} x}\left ( sin^{-1} \frac{x}{a}\right )=\frac{1}{\sqrt{a^{2}-x^{2}}}

\therefore\int \frac{dx}{\sqrt{a^{2}-x^{2}}}=sin^{-1} \frac{x}{a}+c    

-

 

 I_{1} = \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + c

 

Also, I_{1}+ I_{2} = \sin^{-1}\frac{x}{a} + \cos^{-1}\frac{x}{a} + c = \frac{\pi}{2} + c = C 

 


Option 1)

\sin^{-1} \frac{x}{a} +c

Option 2)

\cos^{-1}\frac{x}{a} + c

Option 3)

C

Option 4)

None of these

Posted by

gaurav

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