Question

The value of integral $\int{\dfrac{dx}{\left( \sqrt{1+\sqrt{x}} \right)\sqrt{x-{{x}^{2}}}}}$ is equal to, where C is constant of integration.
\[\begin{align}
  & A.2\sqrt{\dfrac{1+\sqrt{x}}{1-\sqrt{x}}}+C \\
 & B.-2\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}+C \\
 & C.-2\sqrt{\dfrac{1+\sqrt{x}}{1-\sqrt{x}}}+C \\
 & D.-1\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}+C \\
\end{align}\]

Answer 1

Hint: To solve this integral value, we will first solve the square root by substituting value of x as $x={{\cos }^{2}}\theta $. Doing so we will be able to eliminate the square root in the denominator of the given integral. In between we will use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ also we will use ${{\cos }^{2}}x=2{{\cos }^{2}}x-1\Rightarrow \cos x=\dfrac{2{{\cos }^{2}}x-1}{2}$
For the final integral we will use $\int{{{\sec }^{2}}xdx=\tan x+C}$ where C is constant of integral. Because of our assumption we will arrive with $\theta $ as in answer, then we will replace $\theta $ to make answer in form of x using $\tan \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}$

Complete step-by-step answer:
Let the value of integral $\int{\dfrac{dx}{\left( \sqrt{1+\sqrt{x}} \right)\sqrt{x-{{x}^{2}}}}}$ be I.
\[I=\int{\dfrac{dx}{\left( \sqrt{1+\sqrt{x}} \right)\sqrt{x-{{x}^{2}}}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
To solve the under root present in the denominator. Let us assume $x={{\cos }^{2}}\theta $ then differentiating with respect to $\theta $ we get $dx=2\cos \theta \left( -\sin \theta \right)d\theta $ as $\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta $
\[\Rightarrow dx=-2\sin \theta \cos \theta d\theta \]
Substituting this value in equation (i) we get:
\[I=\int{\dfrac{-2\sin \theta \cos \theta }{\left( 1+\sqrt{\left( {{\cos }^{2}}\theta \right)} \right)\sqrt{{{\cos }^{2}}\theta -{{\cos }^{4}}\theta }}}d\theta \]
Replacing $\sqrt{{{\cos }^{2}}\theta }=\cos \theta $ and taking ${{\cos }^{2}}\theta $ common factor $\sqrt{{{\cos }^{2}}\theta -{{\cos }^{4}}\theta }$ we get:
\[I=\int{\dfrac{-2\sin \theta \cos \theta }{\left( 1+\cos \theta \right)\cos \theta \sqrt{1-{{\cos }^{2}}\theta }}}d\theta \]
Now, using trigonometric identity
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \\
 & \Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta } \\
\end{align}\]
Using this in above we get:
\[I=\int{\dfrac{-2\sin \theta \cos \theta }{\left( 1+\cos \theta \right)\left( \sin \theta \right)\left( \cos \theta \right)}}d\theta \]
Cancelling $\sin \theta \cos \theta $ we get:
\[I=\int{\dfrac{-2}{\left( 1+\cos \theta \right)}}d\theta \]
Using the trigonometric identity of cos2x as
\[\begin{align}
  & \cos 2x=2{{\cos }^{2}}x-1 \\
 & \Rightarrow 1+\cos 2x=2{{\cos }^{2}}x \\
\end{align}\]
Replacing 2x = y we get $\dfrac{y}{2}=x$
\[1+\cos y=2{{\cos }^{2}}\dfrac{y}{2}\]
Using this in above replacing by $\theta $ we get:
\[\begin{align}
  & 1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2} \\
 & I=\int{\dfrac{-2}{\left( 2{{\cos }^{2}}\dfrac{\theta }{2} \right)}d\theta } \\
 & I=-\int{\dfrac{1}{{{\cos }^{2}}\dfrac{\theta }{2}}}d\theta \\
\end{align}\]
Now, $\cos x=\dfrac{1}{\sec x}\Rightarrow \sec x=\dfrac{1}{\cos x}$
\[I=-\int{{{\sec }^{2}}}\dfrac{\theta }{2}d\theta \]
And the value of integral of $\int{{{\sec }^{2}}xdx=\tan x}$ using this in value of I.
\[I=\dfrac{-\tan \dfrac{\theta }{2}}{\dfrac{1}{2}}+C\]
Where C is constant of integration.
\[I=-2\tan \dfrac{\theta }{2}+C\]
Now, we need an answer in the form of x so we need $\tan \dfrac{\theta }{2}$ as in form of $\cos \theta $ so that we can calculate the answer in form of x.
Now, we have a trigonometric identity as $\tan \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}$
Using this we get:
\[I=-2\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}+C\]
Now as ${{\cos }^{2}}\theta =x\Rightarrow \cos \theta =\sqrt{x}$
\[I=-2\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}+C\]
Therefore, the value of given integral is $-2\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}+C$

So, the correct answer is “Option C”.

Note: The value of $\tan \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}$ can be obtained by using the trigonometric identity $\cos \theta =\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}$
Cross multiplication of above given:
\[\begin{align}
  & \left( 1+{{\tan }^{2}}\dfrac{\theta }{2} \right)\cos \theta =1-{{\tan }^{2}}\dfrac{\theta }{2} \\
 & \cos \theta +\cos \theta {{\tan }^{2}}\dfrac{\theta }{2}=1-{{\tan }^{2}}\dfrac{\theta }{2} \\
 & \cos \theta +\cos \theta {{\tan }^{2}}\dfrac{\theta }{2}+{{\tan }^{2}}\dfrac{\theta }{2}=1 \\
\end{align}\]
Taking ${{\tan }^{2}}\dfrac{\theta }{2}$ we get:
\[\begin{align}
  & \cos \theta +{{\tan }^{2}}\dfrac{\theta }{2}\left( \cos \theta +1 \right)=1 \\
 & {{\tan }^{2}}\dfrac{\theta }{2}\left( \cos \theta +1 \right)=1-\cos \theta \\
 & {{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{1-\cos \theta }{1+\cos \theta } \\
 & \tan \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }} \\
\end{align}\]
Hence, \[\tan \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\] is obtained.