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# Find the value of the integral $\int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}}$. Choose the correct option.1) $2\left( {\dfrac{{\sqrt x - 1}}{{\sqrt {1 - x} }}} \right) + C$2) $- 2\left( {\dfrac{{\sqrt x + 1}}{{\sqrt {1 - x} }}} \right) + C$3) $2\left( {\dfrac{{\sqrt x - 1}}{{\sqrt {1 + x} }}} \right) + C$4) $2\left( {\dfrac{{\sqrt x + 1}}{{\sqrt {1 + x} }}} \right) + C$

Last updated date: 19th Jun 2024
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Hint: We will start by letting $\sqrt x = \sin \theta$ and differentiate it. After this we will get the value of $dx$ also. So, now we will substitute the values in the original expression given and the integral now becomes in terms of $\sin \theta$. While simplifying the integral, we will use the trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.

Consider the given integral $I = \int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}}$.
We will start by substituting the value of $\sqrt x$ as $\sin \theta$.
Thus, we get,
$\sqrt x = \sin \theta$
Now, square root both the sides of the obtained expression, we get,
$x = {\sin ^2}\theta$
Now, differentiate the above expression, we get,
$dx = 2\sin \theta \cos \theta d\theta$
Now, substitute the obtained values in the given integral,
Thus, we get,
$I = \int {\dfrac{{2\sin \theta \cos \theta d\theta }}{{\left( {1 + \sin \theta } \right)\sqrt {{{\sin }^2}\theta - {{\sin }^4}\theta } }}}$
In the denominator, we can take out ${\sin ^2}\theta$ common from the square root term and take $\sin \theta$ out from the respective term,
Thus, we get,
$\Rightarrow I = \int {\dfrac{{2\sin \theta \cos \theta d\theta }}{{\left( {1 + \sin \theta } \right)\sqrt {{{\sin }^2}\theta \left( {1 - {{\sin }^2}\theta } \right)} }}} \\ \Rightarrow I = \int {\dfrac{{2\sin \theta \cos \theta d\theta }}{{\left( {1 + \sin \theta } \right)\sin \theta \sqrt {1 - {{\sin }^2}\theta } }}} \\$
Now, we will use the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ which gives us the value of $1 - {\sin ^2}\theta = {\cos ^2}\theta$.
Thus, we get,
$\Rightarrow I = \int {\dfrac{{2\sin \theta \cos \theta d\theta }}{{\left( {1 + \sin \theta } \right)\sin \theta \sqrt {{{\cos }^2}\theta } }}} \\ \Rightarrow I = \int {\dfrac{{2\sin \theta \cos \theta d\theta }}{{\left( {1 + \sin \theta } \right)\sin \theta \cos \theta }}} \\ \Rightarrow I = \int {\dfrac{{2d\theta }}{{\left( {1 + \sin \theta } \right)}}} \\$
Next, rationalize by multiplying the numerator and denominator with $1 - \sin \theta$. Thus, we get,
$\Rightarrow I = \int {\dfrac{{2d\theta }}{{\left( {1 + \sin \theta } \right)}} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}} \\ \Rightarrow I = \int {\dfrac{{2\left( {1 - \sin \theta } \right)d\theta }}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}}} \\$
Now, use the property, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ in the denominator.
$\Rightarrow I = \int {\dfrac{{2\left( {1 - \sin \theta } \right)d\theta }}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}}} \\ \Rightarrow I = \int {\dfrac{{2d\theta }}{{\left( {1 + \sin \theta } \right)}} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}} \\ \Rightarrow I = \int {\dfrac{{2\left( {1 - \sin \theta } \right)d\theta }}{{1 - {{\sin }^2}\theta }}} \\$
Now, we will use the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ which gives us the value of $1 - {\sin ^2}\theta = {\cos ^2}\theta$.
$\Rightarrow I = \int {\dfrac{{2\left( {1 - \sin \theta } \right)d\theta }}{{{{\cos }^2}\theta }}}$
Further, simplify the obtained expression,
Thus, we get,
$\Rightarrow I = 2\int {\dfrac{{d\theta }}{{{{\cos }^2}\theta }}} - 2\int {\dfrac{{\sin \theta d\theta }}{{{{\cos }^2}\theta }}}$
We know that, $\dfrac{1}{{{{\cos }^2}\theta }} = {\sec ^2}\theta$ and $\dfrac{{\sin \theta }}{{\cos \theta }} \cdot \dfrac{1}{{\cos \theta }} = \tan \theta \cdot \sec \theta$, use this in the above expression,
Thus, we get,
$\Rightarrow I = 2\int {{{\sec }^2}\theta d\theta } - 2\int {\tan \theta \sec \theta d\theta }$
Here, we know that the direct integral form of ${\sec ^2}\theta$ is $\int {{{\sec }^2}\theta d\theta } = \tan \theta$ and direct integral form of $\tan \theta \sec \theta$ is $\int {\tan \theta \sec \theta d\theta = \sec \theta }$.
Thus, directly apply it in the above integral, we get,
$\Rightarrow I = 2\tan \theta - 2\sec \theta + C$
As we know that, $\sin \theta = \sqrt x$
We can easily find the value of $\cos \theta$ by substituting in the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Thus, we get,
$\Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta } \\ \Rightarrow \cos \theta = \sqrt {1 - x} \\$
Using the values of $\sin \theta = \sqrt x$ and $\cos \theta = \sqrt {1 - x}$, we can find the value of $\tan \theta$.
Thus, we get,
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \\ = \dfrac{{\sqrt x }}{{\sqrt {1 - x} }} \\$
And
$\sec \theta = \dfrac{1}{{\cos \theta }} \\ = \dfrac{1}{{\sqrt {1 - x} }} \\$
Now, substitute the obtained values in the above integral, we get,
$\Rightarrow I = 2\dfrac{{\sqrt x }}{{\sqrt {1 - x} }} - 2\dfrac{1}{{\sqrt {1 - x} }} + C \\ \Rightarrow I = 2\left[ {\dfrac{{\sqrt x - 1}}{{\sqrt {1 - x} }}} \right] + C \\$
Thus, the value of the integral $I = \int {\dfrac{{dx}}{{\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}}$ is $2\left( {\dfrac{{\sqrt x - 1}}{{\sqrt {1 - x} }}} \right) + C$
Hence, option A is correct

Note: Remember to rationalize the expression with $1 - \sin \theta$ at $I = \int {\dfrac{{2d\theta }}{{\left( {1 + \sin \theta } \right)}}}$. Use of trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ is necessary at two or three places while doing the integration. Also, we have used the conversion of $\tan \theta$ and $\sec \theta$ in terms of $\sin \theta$ and $\cos \theta$. It is necessary to remember the direct integral forms of some expressions like in this question we have used $\int {{{\sec }^2}\theta d\theta } = \tan \theta$ and $\int {\tan \theta \sec \theta d\theta = \sec \theta }$. Many students forget to change the expression after integration into x. Keep that in mind after integration we need to take help from trigonometric identities to convert the expression into x as a variable.