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.

& 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{aligned}$$

Answer 1

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

& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \\\ & \Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta } \\\ \end{aligned}$$ Using this in above we get: $$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{aligned} & \cos 2x=2{{\cos }^{2}}x-1 \\\ & \Rightarrow 1+\cos 2x=2{{\cos }^{2}}x \\\ \end{aligned}$$ Replacing 2x = y we get $\dfrac{y}{2}=x$ $$1+\cos y=2{{\cos }^{2}}\dfrac{y}{2}$$ Using this in above replacing by $\theta $ we get: $$\begin{aligned} & 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{aligned}$$ Now, $\cos x=\dfrac{1}{\sec x}\Rightarrow \sec x=\dfrac{1}{\cos x}$ $$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{aligned} & \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{aligned}$$ Taking ${{\tan }^{2}}\dfrac{\theta }{2}$ we get: $$\begin{aligned} & \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{aligned}$$ Hence, $$\tan \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}$$ is obtained.