Question

# $\underset{0}{\overset{a}{\int }}{\mathrm{sin}}^{-1}\sqrt{\frac{x}{a+x}}dx$

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Solution

## $\mathrm{Let},I=\underset{0}{\overset{a}{\int }}{\mathrm{sin}}^{-1}\sqrt{\frac{x}{a+x}}dx\phantom{\rule{0ex}{0ex}}\mathrm{Let},x=a{\mathrm{tan}}^{2}\theta ⇒\theta ={\mathrm{tan}}^{-1}\sqrt{\frac{x}{a}}\phantom{\rule{0ex}{0ex}}\mathrm{When},x\to x;\theta \to 0andx\to a;\theta \to \frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}\mathrm{and}dx=2a\mathrm{tan}\theta se{c}^{2}\theta d\theta \phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}{\mathrm{sin}}^{-1}\sqrt{\frac{a{\mathrm{tan}}^{2}\theta }{a+a{\mathrm{tan}}^{2}\theta }}2a\mathrm{tan}\theta se{c}^{2}\theta d\theta \phantom{\rule{0ex}{0ex}}⇒I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{2a\int }}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\theta \right)\mathrm{tan}\theta se{c}^{2}\theta d\theta \phantom{\rule{0ex}{0ex}}⇒I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{2a\int }}\theta \mathrm{tan}\theta se{c}^{2}\theta d\theta \phantom{\rule{0ex}{0ex}}\mathrm{Let},\mathrm{tan}\theta =t⇒\theta ={\mathrm{tan}}^{-1}t\phantom{\rule{0ex}{0ex}}⇒se{c}^{2}\theta d\theta =dt\phantom{\rule{0ex}{0ex}}\mathrm{when},\theta \to 0;t\to 0and\theta \to \frac{\mathrm{\pi }}{4};t\to 1\phantom{\rule{0ex}{0ex}}\mathrm{Then},I=2a{\int }_{0}^{1}{\mathrm{tan}}^{-1}ttdt\phantom{\rule{0ex}{0ex}}=2a{\int }_{0}^{1}{\mathrm{tan}}^{-1}ttdt\phantom{\rule{0ex}{0ex}}=2a{\left[{\mathrm{tan}}^{-1}t\frac{{t}^{2}}{2}\right]}_{0}^{1}-\frac{2a}{2}{\int }_{0}^{1}\frac{{t}^{2}}{1+{t}^{2}}dt\phantom{\rule{0ex}{0ex}}=2a\left[\frac{\mathrm{\pi }}{4}×\frac{1}{2}-0\right]-a{\int }_{0}^{1}\left[1-\frac{1}{1+{t}^{2}}\right]dt\phantom{\rule{0ex}{0ex}}=2a\left[\frac{\mathrm{\pi }}{8}\right]-a{\left[t-{\mathrm{tan}}^{-1}t\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi a}}{4}-a\left[1-\frac{\pi }{4}\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi a}}{4}-a+\frac{\pi a}{4}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi a}}{2}-a\phantom{\rule{0ex}{0ex}}=a\left(\frac{\pi }{2}-1\right)$

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