Question

Integrate the question in the picture:
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Answer 1

Answer:

$\large\bold\red{ a \: { \tan }^{ - 1} ( \sqrt{ \frac{x}{a} } ) \: { \sec }^{2} \: ( { \tan}^{ - 1} \sqrt{ \frac{x}{a} } ) - \sqrt{ax} + c}$

Step-by-step explanation:

$\displaystyle \int { \sin }^{ - 1} \sqrt{ \frac{x}{a + x} } dx$

Put,

$x = a { \tan }^{2} \alpha$

We get,

$= \displaystyle \int { \sin}^{ - 1} \sqrt{ \frac{a { \tan }^{2} \alpha }{ a(1 + { \tan}^{2} \alpha) } } dx \\ \\ \\ = \displaystyle \int { \sin }^{ - 1} ( \sqrt{ \frac{ { \tan }^{2} \alpha }{ { \sec }^{2} \alpha } } )dx \\ \\ \\ = \displaystyle \int { \sin }^{ - 1} ( \sqrt{ { \sin }^{2} \alpha })dx \\ \\ \\ = \displaystyle \int { \sin }^{ - 1}( \sin \alpha )dx \\ \\ \\ = \displaystyle \int \alpha dx$

Now,

We have,

$x = a { \tan }^{2} \alpha$

Differentiating both sides,

We get,

$= > dx = 2a \tan( \alpha ) { \sec }^{2} \alpha d \alpha$

Therefore,

Substituting the values,

We get,

$= 2a \displaystyle \int \alpha \tan( \alpha ) { \sec}^{2} \alpha d \alpha$

Now,

Put,

$\tan( \alpha ) = t$

Differentiating both the sides,

We get,

$= > { \sec }^{2} \alpha d \alpha = dt$

Also,

$\alpha = { \tan }^{ - 1} t$

Substituting all the values,

We get,

$= 2a\displaystyle \int t \: { \tan }^{ - 1}t \: dt$

Using the ILATE method,

And further solving,

We get,

$= 2a( { \tan}^{ - 1}t \displaystyle \int \: tdt - \displaystyle \int( \frac{d}{dt} { \tan }^{ - 1} t\int \: dt)dt) \\ \\ \\ = 2a( \frac{ {t}^{2} }{2} { \tan }^{ - 1} t - \displaystyle \int \frac{ {t}^{2} }{2} \times \frac{1}{1 + {t}^{2} } dt) \\ \\ \\ = 2a( \frac{ {t}^{2} }{2} { \tan}^{ - 1} t - \displaystyle \int \frac{ {t}^{2} + 1 - 1 }{ {t}^{2} + 1 } dt) \\ \\ \\ = 2a( \frac{ {t}^{2} }{2} { \tan }^{ - 1} t - \frac{1}{2} \displaystyle \int \: dt + \frac{1}{2} \displaystyle \int \frac{1}{1 + {t}^{2} } dt) \\ \\ \\ = 2a (\frac{ {t}^{2} }{2} { \tan }^{ - 1}t - \frac{t}{2} + \frac{1}{2} { \tan }^{ - 1} t + c) \\ \\ \\ = a { \tan}^{ - 1} t( {t}^{2} + 1 )-a t + c$

Substituting the value of t,

We get,

$= a (1 + { \tan }^{2} \alpha ) { \tan }^{ - 1} ( \tan\alpha ) - a \tan( \alpha ) + c \\ \\ \\ = a \: \alpha \: { \sec }^{2} \alpha - a \tan( \alpha ) + c$

Now,

Substituting the value of ,

$\alpha = { \tan }^{ - 1} \sqrt{ \frac{x}{a} }$

We get,

$= a \: { \tan }^{ - 1} ( \sqrt{ \frac{x}{a} } ) { \sec }^{2} ( { \tan}^{ - 1} \sqrt{ \frac{x}{a} } ) - a \tan( { \tan}^{ - 1} \sqrt{ \frac{x}{a} } ) + c \\ \\ \\ = a \: { \tan }^{ - 1} ( \sqrt{ \frac{x}{a} } ) { \sec }^{2} ( { \tan}^{ - 1} \sqrt{ \frac{x}{a} } ) - a \sqrt{ \frac{x}{a} } + c \\ \\ \\ = \large \boxed{ \bold{ a \: { \tan }^{ - 1} ( \sqrt{ \frac{x}{a} } ) \: { \sec }^{2} \: ( { \tan}^{ - 1} \sqrt{ \frac{x}{a} } ) - \sqrt{ax} + c}}$

Where,

c is an integral constant.