Evaluate the following integrals: ^ a _0-1 x a+x dx

Question

Evaluate the following integrals:
$\int^\limits{\text{a}}_0\sin{-1}{\sqrt\frac{\text{x}}{\text{a}+\text{x}}}\text{ dx}$

✓

Answer

Let $\text{I}=\int^\limits{\text{a}}_0\sin{-1}{\sqrt\frac{\text{x}}{\text{a}+\text{x}}}\text{ dx}$
Let $\text{x}=\text{x}\tan^{2}\theta\Rightarrow\theta=\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}$
When, $\text{x}\rightarrow\text{x};\ \theta\rightarrow0$ and $\text{x}\rightarrow\text{a};\ \theta\rightarrow\frac{\pi}{4}$
and $\text{dx}=2\text{a}\tan\theta\ \sec^2\theta\text{ d}\theta$
Then,
$\text{I}=\int^\limits{\frac{\pi}{4}}_0\sin^{-1}\sqrt{\frac{\text{a}\tan^{2}\theta}{\text{a}+\text{a}\tan^{2}\theta}}2\text{a}\tan\theta\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=2\text{a}\int^\limits{\frac{\pi}{4}}_0\sin^{-1}(\sin\theta)\tan\theta\sec^2\theta\text{ d}\theta$
$\text{I}=\int^\limits{\frac{\pi}{4}}_0\theta\tan\theta\sec^2\theta\text{ d}\theta$
Let $\tan\theta=\text{t}\Rightarrow\theta=\tan^{-1}\text{t}$
$\Rightarrow\sec^2\theta\text{ d}\theta=\text{dt}$
When, $\theta\rightarrow0;\text{ t}\rightarrow0$ and $\theta\rightarrow\frac{\pi}{4};\text{ t}\rightarrow1$
Then, $\text{I}=2\text{a}\int^\limits1_0\tan^{-1}\text{t dt}$
$\text{I}=2\text{a}\int^\limits1_0\tan^{-1}\text{t dt}$

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