Question

Find the value of integral
\[\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\sqrt{\cos \theta }}{{{\left( \sqrt{\cos \theta }+\sqrt{\sin \theta } \right)}^{5}}}d\theta }\]

Answer 1

Hint: To solve this question, we will first of all use the definite integral formula stated as
\[\int\limits_{0}^{a}{f\left( x \right)}dx=\int\limits_{0}^{a}{f\left( a-x \right)}dx\] by substituting $x=\dfrac{\pi }{2}$
Then we will add the obtained term to original term to obtain $2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos \theta }+\sqrt{\sin \theta }}{{{\left( \sqrt{\cos \theta }+\sqrt{\sin \theta } \right)}^{5}}}d\theta }$. After cancelling one power of $\sqrt{\cos \theta }+\sqrt{\sin \theta }$ we will make substitution as \[\dfrac{\sqrt{\sin \theta }}{\sqrt{\cos \theta }}=\tan \theta \text{ and }\dfrac{1}{\cos \theta }=\sec \theta \] to get final result. Also, we will use \[\int{\dfrac{1}{{{x}^{k}}}dx=\dfrac{{{x}^{-k+1}}}{-k+1}}\]

Complete step-by-step answer:
Given, $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\sqrt{\cos \theta }}{{{\left( \sqrt{\cos \theta }+\sqrt{\sin \theta } \right)}^{5}}}d\theta }\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
We have a property of definite integral given as below:
\[\int\limits_{0}^{a}{f\left( x \right)}dx=\int\limits_{0}^{a}{f\left( a-x \right)}dx\]
Using this property in equation (i) by taking $a=\dfrac{\pi }{2}$ we get
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\sqrt{\cos \left( \dfrac{\pi }{2}-\theta \right)}}{{{\left( \sqrt{\cos \left( \dfrac{\pi }{2}-\theta \right)}+\sqrt{\sin \left( \dfrac{\pi }{2}-\theta \right)} \right)}^{5}}}d\theta }$
Now, the value of \[\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \] and the value of $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $
Using this in above obtained term we get:
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\sqrt{\sin \theta }}{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{5}}}d\theta }\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$
Now adding equation (i) and equation (ii) we get:
\[\begin{align}
  & 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\sqrt{\cos \theta }}{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{5}}}d\theta }+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\sqrt{\sin \theta }}{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{5}}}d\theta } \\
 & 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3\left( \sqrt{\cos \theta }+\sqrt{\sin \theta } \right)}{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{5}}}d\theta } \\
\end{align}\]
Cancelling $\sqrt{\cos \theta }+\sqrt{\sin \theta }$ from numerator and denominator we get:
\[2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3d\theta }{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{4}}}}\]
Now, dividing by ${{\cos }^{2}}\theta $ both numerator and denominator we get:
\[2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{3}{{{\cos }^{2}}\theta }d\theta }{\dfrac{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{4}}}{{{\cos }^{2}}\theta }}}\]
Using the trigonometric identity $\dfrac{1}{{{\cos }^{2}}\theta }={{\sec }^{2}}\theta \text{ and }\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ in above we have:
\[2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3{{\sec }^{2}}\theta d\theta }{{{\left( \sqrt{\tan \theta }+1 \right)}^{4}}}}\]
Let us assume $\tan \theta ={{t}^{2}}\Rightarrow \sqrt{\tan \theta }=t$ then differentiating both sides with respect to t and using $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ we get:
\[{{\sec }^{2}}\theta d\theta =2tdt\]
When \[\theta =0\Rightarrow \tan \theta =0\Rightarrow \tan \theta ={{t}^{2}}=0\Rightarrow t=0\]
And when \[\theta =\dfrac{\pi }{2}\Rightarrow \tan \dfrac{\pi }{2}={{t}^{2}}\]
Now, value of \[\tan \dfrac{\pi }{2}=\infty \Rightarrow {{t}^{2}}=\infty \Rightarrow t=\infty \]
Applying this substitution in above value of 2I
\[2I=3\int\limits_{0}^{\infty }{\dfrac{2tdt}{{{\left( 1+t \right)}^{4}}}}\]
Cancelling 2 on both sides we get:
And adding and subtracting 1 on numerator to make integral solvable we get:
\[\begin{align}
  & I=3\int\limits_{0}^{\infty }{\dfrac{\left( t+1-1 \right)dt}{{{\left( 1+t \right)}^{4}}}} \\
 & I=3\int\limits_{0}^{\infty }{\left( \dfrac{1+t}{{{\left( 1+t \right)}^{4}}}-\dfrac{1}{{{\left( 1+t \right)}^{4}}} \right)}dt \\
 & I=3\int\limits_{0}^{\infty }{\left( \dfrac{1}{{{\left( 1+t \right)}^{3}}}dt-\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( 1+t \right)}^{4}}}dt} \right)} \\
\end{align}\]
We will now use integration formula:
\[\int{\dfrac{1}{{{x}^{k}}}dx=\dfrac{{{x}^{-k+1}}}{-k+1}}\]
Using this in both terms of above separately to integrate we get:
Let \[P=3\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( 1+t \right)}^{3}}}}dt\text{ and }Q=-3\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( 1+t \right)}^{4}}}}dt\]
Solving P and Q using above formula of integration we get:
\[P=3\left( \dfrac{+1\left( -1 \right)}{2{{\left( 1+t \right)}^{+2}}} \right)_{0}^{\infty }\]
Applying upper and lower limit of integration we get:
\[\begin{align}
  & P=3\left( \dfrac{-1}{\infty }-\dfrac{-1}{2\left( 1+0 \right)} \right) \\
 & \Rightarrow 3\left( 0+\dfrac{1}{2\cdot 1} \right) \\
 & \Rightarrow 3\left( 0+\dfrac{1}{2} \right) \\
 & \Rightarrow 3\left( \dfrac{1}{2} \right)=+\dfrac{3}{2} \\
\end{align}\]
And solving Q similarly, we get:
\[\begin{align}
  & Q=-3\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( 1+t \right)}^{4}}}}dt \\
 & \Rightarrow -3\left( \dfrac{-1}{3{{\left( 1+t \right)}^{3}}} \right)_{0}^{\infty } \\
 & \Rightarrow +3\left( \dfrac{1}{3{{\left( 1+t \right)}^{3}}} \right)_{0}^{\infty } \\
 & \Rightarrow 3\left( \dfrac{1}{\infty }-\dfrac{1}{3\left( 1+0 \right)} \right) \\
 & \Rightarrow 3\left( 0-\dfrac{1}{3} \right) \\
 & \Rightarrow -1 \\
\end{align}\]
Substitute values of P and Q as obtained in I we get:
\[I=\dfrac{+3}{2}+\left( -1 \right)=\dfrac{+3+\left( -2 \right)}{2}=+\dfrac{1}{2}\]
Hence, the value of integral $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos \theta }}{{{\left( \sqrt{\cos \theta }+\sqrt{\sin \theta } \right)}^{5}}}d\theta }=\dfrac{1}{2}=\dfrac{1}{2}$

Note: The possibility of mistake in this question can be at the point where $\tan \theta $ is used as $\dfrac{\sin \theta }{\cos \theta }$
We had \[2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3{{\sec }^{2}}\theta }{\dfrac{{{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{4}}}{{{\cos }^{2}}\theta }}}\]
Now, as we have ${{\cos }^{2}}\theta $ in the denominator part so we will observe how we actually took it inside the root.
We know, $\sqrt{{{\cos }^{2}}\theta }=\cos \theta $
Observe the part ${{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{4}}$
When any term comes out from this bracket then it will be of the form ${{\left( {} \right)}^{4}}$ that is some power of 4. Therefore, when we take any term inside of this bracket then it should be of the form ${{\left( {} \right)}^{\dfrac{1}{4}}}$ that is of power $\dfrac{1}{4}$
So, when we took $\dfrac{1}{{{\cos }^{2}}\theta }$ inside of ${{\left( \sqrt{\sin \theta }+\sqrt{\cos \theta } \right)}^{4}}$ then it becomes of the form \[\dfrac{1}{{{\left( {{\cos }^{2}}\theta \right)}^{\dfrac{1}{4}}}}=\dfrac{1}{{{\cos }^{\dfrac{1}{2}}}\theta }=\dfrac{1}{\cos \theta }\]
\[2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3{{\sec }^{2}}\theta }{{{\left( \dfrac{\sqrt{\sin \theta }}{\sqrt{\cos \theta }}+\dfrac{\sqrt{\cos \theta }}{\sqrt{\cos \theta }} \right)}^{4}}}}d\theta \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{3{{\sec }^{2}}\theta }{{{\left( \sqrt{\tan \theta }+1 \right)}^{4}}}}d\theta \]
is achieved.