Question

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) \[2\left( {\dfrac{{\sqrt x + 1}}{{\sqrt {1 + x} }}} \right) + C\]

Answer 1

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 \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].

Complete step-by-step answer:
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.