Question

For $n>0$ , solution of integral $\int_{0}^{2\pi }{\dfrac{x{{\sin }^{2n}}x}{{{\sin }^{^{2n}}}x+{{\cos }^{2n}}x}dx,}$ is equal to
(a) ${{\pi }^{2}}$
(b) $\dfrac{\pi }{2}$
(c) $2\pi$
(d) $\dfrac{\pi }{4}$

Answer 1

Hint : We have to evaluate the given integral first, let us name it as I. Then, we will use the identity, $\int\limits_{a}^{b}{f\left( x \right)dx=}\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$ , substitute $x$ as $\left( 2\pi -x \right)$ and get a simplified integral. Then, again we will consider this as I and add it to the first integral. Applying the below identities,
$\sin \left( 2\pi -\pi \right)=\operatorname{sinx},$ $\cos \left( 2\pi -x \right)=\operatorname{cosx},$ $\sin \left( \dfrac{\pi }{2}-x \right)=\operatorname{cosx}$ and $\cos \left( \dfrac{\pi }{2}-x \right)=\operatorname{sinx}$ to get the simplify the result. Also, we will be applying another identity $\int\limits_{0}^{2a}{f\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)dx}$ , if $f\left( 2a-x \right)=f\left( x \right)$ .

Complete step-by-step answer :
In the question, we are asked to find the value of integral $\int\limits_{0}^{2\pi }{\dfrac{x{{\sin }^{2n}}x}{{{\sin }^{2\pi }}x+{{\cos }^{2\pi }}x}dx}$ , where $n>0$ .
Now let’s represent the whole integration as ‘I’ so, we get
$I=\int\limits_{0}^{2\pi }{\dfrac{x{{\sin }^{2n}}x}{{{\sin }^{2\pi }}x+{{\cos }^{2\pi }}x}dx}$
Now we will apply identity, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{b}{f\left( a+b-x \right)}dx$
Here we will substitute $x$ as $\left( 2\pi -x \right)$ , so we get,
$I=\int\limits_{0}^{2\pi }{\dfrac{\left( 2\pi -x \right){{\sin }^{2n}}\left( 2\pi -x \right)}{{{\sin }^{2\pi }}\left( 2\pi -x \right)+{{\cos }^{2\pi }}\left( 2\pi -x \right)}}dx$
We know that $\sin \left( 2\pi -\theta \right)=\sin \theta$ and $\cos \left( 2\pi -\theta \right)=\cos \theta$ , so we get,
$I=\int\limits_{0}^{2\pi }{\dfrac{\left( 2\pi -x \right){{\sin }^{2n}}}{{{\sin }^{2\pi }}+{{\cos }^{2\pi }}x}dx}$
Now we will add this expression of ‘I’ with original expression
$I+I=\int\limits_{0}^{2\pi }{\dfrac{\left( 2\pi -x \right){{\sin }^{2n}}x+{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}dx$
$2I=2\pi \int\limits_{0}^{2\pi }{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}dx$
$I=\pi \int\limits_{0}^{2\pi }{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}dx$
Now we can write ‘I’ as, $I=\pi \int\limits_{0}^{2\pi }{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}dx$ because $I=2\pi \int\limits_{0}^{\pi }{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$ , by using identity $\int\limits_{0}^{2a}{f\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)dx}$ , if $f\left( 2a-x \right)=f\left( x \right)$ , where $f\left( x \right)$ is $\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}$ and ‘a’ is ‘ $\pi$ ’.
Now we will further write ‘I’ as,
$I=2\pi \int\limits_{0}^{2\left( \dfrac{\pi }{2} \right)}{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$
By using identity $\int\limits_{0}^{2a}{f\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)dx}$
If $f\left( 2a-x \right)=f\left( x \right)$ where $f\left( x \right)$ is $\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}$ and ‘ $a$ ’ is $\dfrac{\pi }{2}$ .
So, we get $I=4\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}dx$
Now we will once again use the identity
$\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$
But here we will put x as $\left( \dfrac{\pi }{2}-x \right)$ so we get,
$I=4\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2n}}\left( \dfrac{\pi }{2}-x \right)}{{{\sin }^{2n}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{2n}}\left( \dfrac{\pi }{2}-x \right)}dx}$
Then we will apply identity
$\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta ,\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \dfrac{\pi }{4}\theta$
So we can write as, $I=4\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$
But we also know that $I=4\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$
So we can write,
$I+I=4\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2n}}x+{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}dx}$
$\begin{aligned} & 2I=4\pi \int\limits_{0}^{\dfrac{\pi }{2}}{dx} \\\ & 2I=4\pi \left[ x \right]_{0}^{\dfrac{\pi }{2}} \\\ & 2I=4\pi \left[ \dfrac{\pi }{2}-0 \right] \\\ \end{aligned}$
Hence $2I=4\pi \left( \dfrac{\pi }{2} \right)\Rightarrow I={{\pi }^{2}}$
So the value of ‘I’ is ${{\pi }^{2}}$ .
The correct option is (a).

Note : Students should be careful while applying identities and also about calculations too. This is because any mistake can end up being a solution getting wrong. The step $I+I=\int\limits_{0}^{2\pi }{\dfrac{\left( 2\pi -x \right){{\sin }^{2n}}x+{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}dx$ is where most of them tend to make the mistake of forgetting to add I + I on the LHS. So, this makes the whole solution wrong. Some students even go wrong with the identity and take it as $\sin \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta ,\cos \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta$ , but whenever we use angle $\dfrac{\pi }{2}$ , the function changes.