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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 58 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 58  Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{2}\left ( x\cos ^{-1}x-\sqrt{1-x^{2}} \right )+c

Hint: Let x=\cos \Theta

Given: \int \tan ^{-1}\sqrt{\frac{1-x}{1+x}}dx

Solution: Let x=\cos \Theta

                dx=-\sin \Theta \: d\: \Theta

\therefore \int \tan ^{-1}\sqrt{\frac{1-x}{1+x}}dx=\int \tan ^{-1}\sqrt{\frac{1-\cos \Theta }{1+\cos \Theta }}\left ( -\sin \Theta \: d\Theta \right )

                                          \begin{aligned} &=-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin \theta d \theta \\ &=-\int \tan ^{-1} \tan \frac{\theta}{2} \sin \theta d \theta \\ &=-\frac{1}{2} \int \theta \sin \theta d \theta \\ &=-\frac{1}{2}\left[\theta(-\cos \theta)-\int 1(-\cos \theta) d \theta\right] \end{aligned}

Applying by parts

=\frac{1}{2}\left [ -\Theta \cos \Theta +\sin \Theta \right ]

\begin{aligned} &=\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta \\ &=\frac{1}{2} \cos ^{-1} x x-\frac{1}{2} \sqrt{1-x^{2}}+c \\ &=\frac{x}{2} \cos ^{-1} x-\frac{1}{2} \sqrt{1-x^{2}}+c \\ &=\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^{2}}\right)+c \end{aligned}

 

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