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  • Integrate the functions in Exercises 1 to 24. 23

Integrate the functions in Exercises 1 to 24.

    Q23.    \tan^{-1}\sqrt{\frac{1-x}{1+x}}

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We have                    

                                              I\ =\ \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}

Let us assume that  :                   x\ =\ \cos 2\Theta

                                         Differentiating wrt x,

                                                dx\ =\ -2 \sin 2\Theta\ d\Theta

Substituting this in the original equation, we get 

                                              \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\ =\ \int \tan^{-1}\sqrt{\frac{1-cos2\Theta }{1+cos2\Theta }}\times -2\sin 2\Theta \ d\Theta

or                                                                                  =\ -2\int \tan^{-1} (\frac{sin\Theta }{cos\Theta })\times \sin 2\Theta \ d\Theta

or                                                                                   =\ -2\int \Theta \sin 2\Theta \ d\Theta

Using integration by parts, we get 

                                                                                     =\ -2\left ( \Theta \int \sin 2\Theta \ d\Theta\ - \int \frac{d\Theta }{d\Theta } \int \sin 2\Theta \ d\Theta\ \right )

or                                                                                  =\ -2\left ( \Theta \left ( \frac{-\cos 2\Theta }{2} \right ) - \int 1.\frac{-\cos 2\Theta }{2} \ d\Theta\ \right )

or                                                                                   =\ -2\left ( \frac{-\Theta \cos 2\Theta }{2}+ \frac{\sin 2\Theta }{4} \right )

Putting all the assumed values back in the expression,

                                                                                       =\ -2\left ( -\frac{1}{2}\left ( \frac{1}{2} \cos^{-1} x \right )+ \frac{\sqrt{1-x^2} }{4} \right )

or                                                                                    =\ \frac{1}{2}\left ( x \cos^{-1} x\ -\ \sqrt{1-x^2} \right )\ +\ C 

Posted by

Devendra Khairwa

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