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{align}
  & 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{align} $
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.