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

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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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