Question

The integral \[\int{\dfrac{dx}{\left( 1+\sqrt{x} \right)\sqrt{x-{{x}^{2}}}}}\] is equal to (where c is a constant of integration):
\[\begin{align}
  & \left( \text{a} \right)\text{ }-2\sqrt{\dfrac{1+\sqrt{x}}{1-\sqrt{x}}}+c \\
 & \left( \text{b} \right)\text{ }-2\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}+c \\
 & \left( \text{c} \right)\text{ }2\left[ \dfrac{\sqrt{x}-1}{\sqrt{1-x}} \right]+c \\
 & \left( \text{d} \right)\text{ }-\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}+c \\
\end{align}\]

Answer 1

Hint: Put \[\sqrt{x}=\sin \theta \], squaring on both sides we will get \[x={{\sin }^{2}}\theta \], substitute these values in the integral. Use basic integral formulae to solve the integral. After getting the simplified integral, replace \[\sin \theta \] by \[\sqrt{x}\].

Complete step-by-step solution:
Given the integral is,
\[\int{\dfrac{1}{\left( 1+\sqrt{x} \right)\sqrt{x-{{x}^{2}}}}}.dx-(1)\]
Let us consider,
\[\sqrt{x}=\sin \theta \].
Now squaring on both sides we get,
\[x={{\sin }^{2}}\theta \]
Differentiating it, we get,
\[dx=2\sin \theta \cos \theta .d\theta \]
Putting the value of \[\sqrt{x},x\] and \[dx\] in equation (1), we get
\[I=\int{\dfrac{2\sin \theta \cos \theta d\theta }{\left( 1+\sin \theta \right)\sqrt{{{\sin }^{2}}\theta -{{\sin }^{4}}\theta }}}\]
Take \[\left( {{\sin }^{2}}\theta \right)\] common from the square root, we get
\[I=\int{\dfrac{2\sin \theta \cos \theta .d\theta }{\left( 1+\sin \theta \right)\sin \theta \sqrt{1-{{\sin }^{2}}\theta }}}\]
We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \], so we get
\[I=\int{\dfrac{2\sin \theta \cos \theta .d\theta }{\left( 1+\sin \theta \right)\sin \theta \cos \theta }}\]
Cancel out \[\left( \sin \theta .\cos \theta \right)\] from numerator & denominator.
\[=\int{\dfrac{2.d\theta }{1+\sin \theta }=2\int{\dfrac{d\theta }{1+\sin \theta }}}\]
Multiply \[\left( 1-\sin \theta \right)\] on numerator and denominator.
\[=2\int{\dfrac{\left( 1-\sin \theta \right)d\theta }{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}}\]
We know \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]
\[\begin{align}
  & \Rightarrow I=2\int{\dfrac{\left( 1-\sin \theta \right)d}{1-{{\sin }^{2}}\theta }}=2\int{\dfrac{1-\sin \theta }{{{\cos }^{2}}\theta }d\theta } \\
 & \Rightarrow 2\int{\left( \dfrac{1}{{{\cos }^{2}}\theta }-\dfrac{\sin \theta }{{{\cos }^{2}}\theta } \right)d\theta } \\
\end{align}\]
We know, \[\dfrac{1}{\cos \theta }=\sec \theta \Rightarrow \dfrac{1}{{{\cos }^{2}}\theta }={{\sec }^{2}}\theta ,\dfrac{\sin \theta }{\cos \theta }=\tan \theta \], so above equation can be written as,
\[\begin{align}
  & =2\int{\left( {{\sec }^{2}}\theta -\tan \theta \sec \theta \right)d\theta } \\
 & =2\int{{{\sec }^{2}}\theta .d\theta }-2\int{\tan \theta \sec \theta .d\theta } \\
\end{align}\]
We know, \[\int{{{\sec }^{2}}\theta .d\theta }=\tan \theta +c,\int{\tan \theta \sec \theta }.d\theta =\sec \theta +c\], so above equation can be written as
\[\therefore I=2\left[ \tan \theta -\sec \theta \right]+c-(2)\]
We took \[\sin \theta =\sqrt{x}\]and \[x={{\sin }^{2}}\theta \]
So we get,
\[\begin{align}
  & \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-x} \\
 & \tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{\sqrt{x}}{\sqrt{1-x}} \\
 & \sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\sqrt{1-x}} \\
\end{align}\]
So the equation (2) can be written as,
\[\begin{align}
  & I=2\left[ \dfrac{\sqrt{x}}{\sqrt{1-x}}-\dfrac{1}{\sqrt{1-x}} \right]+c \\
 & I=2\left[ \dfrac{\sqrt{x}-1}{\sqrt{1-x}} \right]+c \\
\end{align}\]
Hence, the correct option is (d).

Note: By putting \[\sqrt{x}=\cos \theta \], we cannot find the required solution. Therefore, put \[\sqrt{x}=\sin \theta \] to solve the integral. You should remember the basic integration formulas, which are required for the solution. You can solve most of the steps using basic identities and functions. However, to get the final answer you need integration formulas.