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

Question

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

✓

Answer

Let $\text{I}=\int\sin^{-1}\sqrt{\frac{\text{x}}{\text{a+x}}}\text{dx}$
Let $\text{x}=\text{a}\tan^2\theta$
$\text{dx}=2\text{a}\tan\theta\sec^2\theta\text{d}\theta$
$\text{I}=\int\Big(\sin^{-1}\sqrt{\frac{\text{a}\tan^2\theta}{\text{a+a}\tan^2\theta}}\Big)(2\text{a}\tan\theta\sec^2\theta)\text{d}\theta$
$=\int\Big(\sin^{-1}\sqrt{\frac{\tan^2\theta}{\sec^2\theta}}\Big)(2\text{a}\tan\theta\sec^2\theta)\text{d}\theta$
$=\int\sin^{-1}(\sin\theta)(2\text{a}\tan\theta\sec^2\theta)\text{d}\theta$
$=\int2\theta\text{a}\tan\theta\sec^2\theta\text{d}\theta$
$=2\text{a}\int\theta(\tan\theta\sec^2\theta)\text{d}\theta$
$=2\text{a}\big[\theta\int\tan\theta\sec^2\theta\text{d}\theta-\int(\int\tan\theta\sec^2\theta\text{d}\theta)\text{d}\theta\Big]$
$=2\text{a}\Big[\theta\frac{\tan^2\theta}{2}-\int\frac{\tan^2\theta}{2}\text{d}\theta\Big]$
$=\text{a}\theta\tan^2\theta-\frac{2\text{a}}{2}\int(\sec^2\theta-1)\text{d}\theta$
$=\text{a}\theta\tan^2\theta-\text{a}\tan\theta+\text{a}\theta+\text{C}$
$=\text{a}\Big(\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}\Big)\frac{\text{x}}{\text{a}}-\text{a}\sqrt{\frac{\text{x}}{\text{a}}}+\text{a}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}+\text{C}$
$\text{I}=\text{x}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\text{ax}}+\text{a}\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}+\text{C}$

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