Find: $\displaystyle \int \sin ^{-1} \sqrt{\frac{x}{\mathrm{a}+x}} \mathrm{~d} x$.

Put $\displaystyle x=\mathrm{a} \tan ^{2} \theta, d x=2 \mathrm{a} \tan \theta \sec ^{2} \theta d \theta$ \[\begin{aligned} \int \sin ^{-1} \sqrt{\frac{x}{\mathrm{a}+x}} d x & =2 \mathrm{a} \int \sin ^{-1} \sqrt{\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}}\left(\tan \theta \cdot \sec ^{2} \theta\right) d \theta \\ & =2 \mathrm{a} \int \theta \cdot\left(\tan \theta \sec ^{2} \theta\right) d \theta \\ & =2 \mathrm{a}\left[\theta \frac{(\tan \theta)^{2}}{2}-\frac{1}{2} \int(\tan \theta)^{2} d \theta\right] \\ & =\mathrm{a}\left[\theta \tan ^{2} \theta-(\tan \theta-\theta)\right]+C \\ & =\mathrm{a}\left[\tan ^{-1}\left(\sqrt{\frac{x}{\mathrm{a}}}\right) \cdot \frac{x}{\mathrm{a}}-\sqrt{\frac{x}{\mathrm{a}}}+\tan ^{-1}\left(\sqrt{\frac{x}{\mathrm{a}}}\right)\right]+C \\ & \text { or }(\mathrm{a}+x) \tan ^{-1}\left(\sqrt{\frac{x}{\mathrm{a}}}\right)-\sqrt{\mathrm{a} x}+C \end{aligned} \]