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The integral \int sec^{\frac{2}{3}}x\; cosec^{\frac{4}{3}}x\; dx is equal to :

(Here C is a constant of integration)

  • Option 1)

     -3\; tan^{\frac{-1}{3}}x+C            

  • Option 2)

    -\; \frac{3}{4}\; tan^{\frac{-4}{3}}x+C

  • Option 3)

    -3\; cot^{\frac{-1}{3}}x+C

  • Option 4)

     3\; tan^{\frac{-1}{3}}x+C

Answers (1)

best_answer

\int sec^{\frac{2}{3}}x\; cosec^{\frac{4}{3}}x\; dx

           cosec\left ( x \right )=\frac{sec\left ( x \right )}{tan\left ( x \right )}

                 =\int sec^{2}\left ( x \right )\cdot \frac{1}{tan^{\frac{4}{3}}\left ( x \right )}dx

                put,  tan\left ( x \right )=t

                  sec^{2}\left ( x \right )dx=dt

                    =\int \frac{1}{t^{^{\frac{4}{3}}}}dt

                   =\int -\frac{3}{t^{\frac{1}{3}}}+C

                   =-3\left ( cot^{\frac{1}{3}}\left ( x \right ) \right )

                       or

                 =\frac{-3}{tan^{\frac{1}{3}}\left ( x \right )}


Option 1)

 -3\; tan^{\frac{-1}{3}}x+C            

Option 2)

-\; \frac{3}{4}\; tan^{\frac{-4}{3}}x+C

Option 3)

-3\; cot^{\frac{-1}{3}}x+C

Option 4)

 3\; tan^{\frac{-1}{3}}x+C

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