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  • Stuck here, help me understand: - Integral Calculus - BITSAT

\int \frac{x^{2}+ \cos^2x}{(1+x^{2}) \sin^{2}x}.dx

is equal to?

  • Option 1)

    \tan^{-1}x+\cot x+ c

  • Option 2)

    \tan^{-1}x-\cot x+ c

  • Option 3)

    \cot^{-1} x- \tan^x+ c

  • Option 4)

    -\tan^{-1} x- \cot x+ c

 

Answers (1)

best_answer

\int \frac{x^{2}+\cos ^{2}x}{(1+x^{2})\sin ^{2}x} \: dx

=\int \frac{x^{2}+1-\sin ^{2}x}{(1+x^{2})\sin ^{2}x} \: dx

=\int \frac{(1+x^{2})}{(1+x^{2})\sin ^{2}x}dx-\int \frac{1}{(1+x^{2})}dx

\int cosec^{2}xdx-\int \frac{1}{1+x^{2}}dx

=-cotx-tan^{-1}x+c

 


Option 1)

\tan^{-1}x+\cot x+ c

This option is incorrect

Option 2)

\tan^{-1}x-\cot x+ c

This option is incorrect

Option 3)

\cot^{-1} x- \tan^x+ c

This option is incorrect

Option 4)

-\tan^{-1} x- \cot x+ c

This option is correct

Posted by

Aadil

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