Question

Evaluate the following integral: - $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$.

Answer 1

Hint: Assume the given integral as I. Now, write the denominator of the given function as ${{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x$ and simplify. Use the trigonometric identities $2\sin x\cos x=\sin 2x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to further simplify the denominator. Substitute $\cos 2x=k$ and differentiate both the sides using the formula $\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ to find the value of $dx$ in terms of $dk$. Substitute all the values in the given integral and change the limits according to the assumptions. Finally, use the formula $\int{\dfrac{1}{1+{{k}^{2}}}dk}={{\tan }^{-1}}k$ and substitute the limits to get the answer.

Complete step by step answer:
Here we have been provided with the definite integral $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$ and we are asked to find its value. Assuming the integral as I we have,
$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$
Now, let us simplify the function present inside the integral sign, so we can write ${{\sin }^{4}}x+{{\cos }^{4}}x$ as: -
\[\begin{align}
  & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x \\
 & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2\times \dfrac{{{\left( 2\sin x\cos x \right)}^{2}}}{4} \\
\end{align}\]
Using the trigonometric identities $2\sin x\cos x=\sin 2x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ we get,
\[\begin{align}
  & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x={{\left( 1 \right)}^{2}}-\dfrac{{{\left( \sin 2x \right)}^{2}}}{2} \\
 & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x=1-\dfrac{{{\sin }^{2}}2x}{2} \\
\end{align}\]
Further using the trigonometric identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ we get,
\[\begin{align}
  & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x=1-\dfrac{1-{{\cos }^{2}}2x}{2} \\
 & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x=\dfrac{1+{{\cos }^{2}}2x}{2} \\
\end{align}\]
Substituting the above simplified expression in the integral I we get,
$\begin{align}
  & \Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{\left( \dfrac{1+{{\cos }^{2}}2x}{2} \right)}dx} \\
 & \Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2\sin 2x}{1+{{\cos }^{2}}2x}dx} \\
\end{align}$
Substituting $\cos 2x=k$ and differentiating both the sides using the formula $\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ to find the value of $dx$ in terms of $dk$, we get,
$\Rightarrow \dfrac{d\left( \cos 2x \right)}{dx}=-2\sin 2x$
Now, according to the assumptions we have to change the upper and the lower limit values. At x = 0 we will have $k=\cos 0=1$ and at $x=\dfrac{\pi }{2}$ we will have$k=\cos \pi =-1$. Therefore, the integral with the new limits will be given as: -
$\Rightarrow I=-\int_{1}^{-1}{\dfrac{1}{1+{{k}^{2}}}dk}$
Using the formula $\int{\dfrac{1}{1+{{k}^{2}}}dk}={{\tan }^{-1}}k$ and substituting the suitable limits we get,
$\begin{align}
  & \Rightarrow I=-\left[ {{\tan }^{-1}}k \right]_{1}^{-1} \\
 & \Rightarrow I=-\left[ {{\tan }^{-1}}\left( -1 \right)-{{\tan }^{-1}}\left( 1 \right) \right] \\
 & \Rightarrow I={{\tan }^{-1}}\left( 1 \right)-{{\tan }^{-1}}\left( -1 \right) \\
\end{align}$
We knew that ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ and ${{\tan }^{-1}}\left( -1 \right)=-\dfrac{\pi }{4}$, so we get,
$\begin{align}
  & \Rightarrow I=\dfrac{\pi }{4}-\left( -\dfrac{\pi }{4} \right) \\
 & \Rightarrow I=\dfrac{\pi }{4}+\dfrac{\pi }{4} \\
 & \therefore I=\dfrac{\pi }{2} \\
\end{align}$
Hence, the value of the given definite integral is $\dfrac{\pi }{2}$.

Note: Note that here you can use one of the properties of the definite integral given as$\int_{0}^{a}{f\left( x \right)dx}=\int_{0}^{a}{f\left( a-x \right)dx}$. On using this property you will get the function will satisfy the relation $f\left( x \right)=f\left( a-x \right)$ and due to this the integral will become $2\int_{0}^{\dfrac{a}{2}}{f\left( x \right)dx}$. However, after this you have to apply the same approach as we have used above to get the answer, i.e. the method of substitution. Do not forget to change the limits while performing integration by substitution.