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  • Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 59 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 59 Maths Textbook Solution.

Answers (1)

Answer: a\left [ \left ( \frac{x}{a}+1 \right )\tan ^{-1}\sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}} \right ]+c

Hint: Put x=a\tan ^{2}\Theta

Given:\int \sin ^{-1}\left ( \sqrt{\frac{x}{a+x}} \right )dx

Solution:\int \sin ^{-1}\left ( \sqrt{\frac{x}{a+x}} \right )dx

          Put x=a\tan ^{2}\theta \Rightarrow dx=2a\tan \theta \sec ^{2}\theta d\theta

              \int \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right) d x=\int \sin ^{-1}\left(\sqrt{\frac{a \tan ^{2} \theta}{a+a \tan ^{2} \theta}}\right) 2 a \tan \theta \sec ^{2} \theta d \theta

                                                           \begin{aligned} &=2 a \int \sin ^{-1}(\sin \theta) \tan \theta \sec ^{2} \theta d \theta \\ &=2 a \int \theta \tan \theta \sec ^{2} \theta d \theta \end{aligned}

Let \tan \Theta =t\Rightarrow \sec ^{2}\Theta d\Theta =dt

      \begin{aligned} &\Rightarrow 2 a \int t \tan ^{-1} t d t \\ &\Rightarrow 2 a\left[\tan ^{-1} t \int t d t-\int\left(\frac{1}{1+t^{2}}\right) \frac{t^{2}}{2} d t\right] \\ &\Rightarrow 2 a\left[\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2} \int\left(1-\frac{1}{1+t^{2}}\right) d t\right] \\ &\Rightarrow a\left[t^{2} \tan ^{-1} t-t+\sin ^{-1} t\right]+c \\ &\Rightarrow a\left[\tan ^{2} \theta \tan \theta-\tan \theta+\tan \theta\right]+c \\ &\Rightarrow a\left[\frac{x}{a} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}+\sqrt{\frac{x}{a}}\right]+c \end{aligned}

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