Evaluate the given indefinite integral: $ \int{\sqrt{\dfrac{a-x}{a+x}}dx} $ ?
(a) $ a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\sqrt{{{a}^{2}}-{{x}^{2}}}+c $
(b) $ a{{\cos }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}+{{x}^{2}}}+c $
(c) $ {{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+c $
(d) $ a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+c $ .

Hint: We start solving the problem by assuming $ x=a\cos \theta $ and finding $ dx $ in terms of $ d\theta $ . We then substitute these in the integrand and make the necessary calculations to proceed through the problem. We then make use of the facts $ \int{pdx}=px+c $ and $ \int{p\cos xdx}=p\sin x+c $ to get the value of integral in independent variable $ \theta $ . We then convert $ \theta $ to x from $ x=a\cos \theta $ and make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the given indefinite integral: $ \int{\sqrt{\dfrac{a-x}{a+x}}dx} $ .
Let us assume $ I=\int{\sqrt{\dfrac{a-x}{a+x}}dx} $ ---(1).
Let us assume $ x=a\cos \theta $ ---(2).
Let us apply differential on both sides of equation (2).
 $ \Rightarrow d\left( x \right)=d\left( a\cos \theta \right) $ .
 $ \Rightarrow dx=-a\sin \theta d\theta $ ---(3).
Let us substitute equations (2) and (3) in equation (1).
 $ \Rightarrow I=\int{\sqrt{\dfrac{a-a\cos \theta }{a+a\cos \theta }}\times \left( -a\sin \theta \right)d\theta } $ .
 $ \Rightarrow I=\int{\sqrt{\dfrac{a\left( 1-\cos \theta \right)}{a\left( 1+\cos \theta \right)}}\times \left( -a\sin \theta \right)d\theta } $ .
 $ \Rightarrow I=\int{\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\times \left( -a\sin \theta \right)d\theta } $ .
We know that $ 1-\cos \theta =2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right) $ and $ 1+\cos \theta =2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right) $ .
 $ \Rightarrow I=\int{\sqrt{\dfrac{2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)}{2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)}}\times \left( -a\sin \theta \right)d\theta } $ .
 $ \Rightarrow I=\int{\sqrt{\dfrac{{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)}{{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)}}\times \left( -a\sin \theta \right)d\theta } $ .
 $ \Rightarrow I=\int{\dfrac{\sin \left( \dfrac{\theta }{2} \right)}{\cos \left( \dfrac{\theta }{2} \right)}\times \left( -a\sin \theta \right)d\theta } $ .
We know that $ \sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right) $ .
 $ \Rightarrow I=\int{\dfrac{\sin \left( \dfrac{\theta }{2} \right)}{\cos \left( \dfrac{\theta }{2} \right)}\times \left( -2a\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right) \right)d\theta } $ .
 $ \Rightarrow I=-a\int{2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)d\theta } $ .
We know that $ 2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)=1-\cos \theta $ .
 $ \Rightarrow I=-a\int{\left( 1-\cos \theta \right)d\theta } $ .
 $ \Rightarrow I=\int{-ad\theta }+\int{a\cos \theta d\theta } $ .
We know that $ \int{pdx}=px+c $ and $ \int{p\cos xdx}=p\sin x+c $ .
 $ \Rightarrow I=-a\theta +a\sin \theta +c $ ---(4).
But we have $ x=a\cos \theta $ .
 $ \Rightarrow \dfrac{x}{a}=\cos \theta $ .
 $ \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{x}{a} \right) $ ---(5).
 $ \Rightarrow \dfrac{x}{a}=\cos \theta $ .
 $ \Rightarrow {{\cos }^{2}}\theta =\dfrac{{{x}^{2}}}{{{a}^{2}}} $ .
 $ \Rightarrow 1-{{\sin }^{2}}\theta =\dfrac{{{x}^{2}}}{{{a}^{2}}} $ .
 $ \Rightarrow 1-\dfrac{{{x}^{2}}}{{{a}^{2}}}={{\sin }^{2}}\theta $ .
 $ \Rightarrow {{\sin }^{2}}\theta =\dfrac{{{a}^{2}}-{{x}^{2}}}{{{a}^{2}}} $ .
 $ \Rightarrow \sin \theta =\dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} $ ---(6).
Let us substitute equations (5) and (6) in equation (4).
 $ \Rightarrow I=-a{{\cos }^{-1}}\left( \dfrac{x}{a} \right)+a\times \left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)+c $ .
 $ \Rightarrow I=-a{{\cos }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+c $ .
 $ \Rightarrow I=-a\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{x}{a} \right) \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+c $ .
 $ \Rightarrow I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+c-\dfrac{a\pi }{2} $ .
Let us assume $ c-\dfrac{a\pi }{2}=C $ , as both are constants.
 $ \Rightarrow I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+C $ .
So, we have found the result of the given definite integral as $ a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+C $ .
 $ \therefore, $ The correct option for the given problem is (d).

Note:
 We can also solve this problem as shown below:
We have $ I=\int{\sqrt{\dfrac{a-x}{a+x}}dx} $ . Let us multiply the numerator and denominator inside the square root with $ a-x $ .
 $ \Rightarrow I=\int{\sqrt{\dfrac{a-x}{a+x}\times \dfrac{a-x}{a-x}}dx} $ .
 $ \Rightarrow I=\int{\sqrt{\dfrac{{{\left( a-x \right)}^{2}}}{{{a}^{2}}-{{x}^{2}}}}dx} $ .
 $ \Rightarrow I=\int{\dfrac{a-x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} $ .
 $ \Rightarrow I=\int{\dfrac{a}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}-\int{\dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} $ .
 $ \Rightarrow I={{I}_{1}}-{{I}_{2}} $ ---(7).
Let us first solve for $ {{I}_{1}} $ .
 $ \Rightarrow {{I}_{1}}=\int{\dfrac{a}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} $ ---(8).
Let us assume $ x=a\sin \theta $ ---(9).
Let us apply a differential on both sides of the equation (9).
 $ \Rightarrow d\left( x \right)=d\left( a\sin \theta \right) $ .
 $ \Rightarrow dx=a\cos \theta d\theta $ ---(10).
\[\Rightarrow {{I}_{1}}=\int{\dfrac{a}{\sqrt{{{a}^{2}}-{{\left( a\sin \theta \right)}^{2}}}}\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{a}{\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta }}\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{a}{\sqrt{{{a}^{2}}{{\cos }^{2}}\theta }}\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{a}{a\cos \theta }\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{1}}=\int{ad\theta }\].
\[\Rightarrow {{I}_{1}}=a\theta +{{c}_{1}}\].
\[\Rightarrow {{I}_{1}}=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+{{c}_{1}}\] ---(11).
Let us first solve for $ {{I}_{2}} $ .
 $ \Rightarrow {{I}_{2}}=\int{\dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} $ ---(12).
Let us assume $ x=a\sin \theta $ ---(13).
Let us apply a differential on both sides of the equation (13).
 $ \Rightarrow d\left( x \right)=d\left( a\sin \theta \right) $ .
 $ \Rightarrow dx=a\cos \theta d\theta $ ---(14).
\[\Rightarrow {{I}_{2}}=\int{\dfrac{\left( a\sin \theta \right)}{\sqrt{{{a}^{2}}-{{\left( a\sin \theta \right)}^{2}}}}\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{2}}=\int{\dfrac{\left( a\sin \theta \right)}{\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta }}\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{2}}=\int{\dfrac{\left( a\sin \theta \right)}{\sqrt{{{a}^{2}}{{\cos }^{2}}\theta }}\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{2}}=\int{\dfrac{\left( a\sin \theta \right)}{a\cos \theta }\left( a\cos \theta \right)d\theta }\].
\[\Rightarrow {{I}_{2}}=\int{\left( a\sin \theta \right)d\theta }\].
\[\Rightarrow {{I}_{2}}=-a\cos \theta +{{c}_{2}}\].
\[\Rightarrow {{I}_{2}}=-a\sqrt{{{\cos }^{2}}\theta }+{{c}_{2}}\].
\[\Rightarrow {{I}_{2}}=-a\sqrt{1-{{\sin }^{2}}\theta }+{{c}_{2}}\].
\[\Rightarrow {{I}_{2}}=-a\sqrt{1-{{\left( \dfrac{x}{a} \right)}^{2}}}+{{c}_{2}}\].
\[\Rightarrow {{I}_{2}}=-a\sqrt{\dfrac{{{a}^{2}}-{{x}^{2}}}{{{a}^{2}}}}+{{c}_{2}}\].
\[\Rightarrow {{I}_{2}}=-\sqrt{{{a}^{2}}-{{x}^{2}}}+{{c}_{2}}\] ---(15).
Let us substitute equations (11) and (15) in equation (7).
 $ \Rightarrow I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+{{c}_{1}}-\left( -\sqrt{{{a}^{2}}-{{x}^{2}}}+{{c}_{2}} \right) $ .
 $ \Rightarrow I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+{{c}_{1}}+{{c}_{2}} $ .
Let us assume $ {{c}_{1}}+{{c}_{2}}=c $ .
 $ \Rightarrow I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+\sqrt{{{a}^{2}}-{{x}^{2}}}+c $ .