Question

Find the value of the following integral $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$

Answer 1

Hint: To solve this integral we will first use the property of definite integral which says $\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)}$ hence using this we can further simplify the equation and get the value of integral as \[\dfrac{\pi }{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}\]. Now here we will use ${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab$ to further simplify the equation and convert the terms to $\sin 2x$and $\cos 2x$ with the formula $\sin 2x=2\sin x\cos x$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ after this we will substitute $\cos 2x=t$ and solve the definite integral.

Complete step-by-step answer:
The given integral is $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$
Now by the property of definite integral we know that $\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)}$
Hence we get.
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( 0+\dfrac{\pi }{2}-x \right)\sin \left( 0+\dfrac{\pi }{2}-x \right)\cos \left( 0+\dfrac{\pi }{2}-x \right)}{{{\sin }^{4}}\left( 0+\dfrac{\pi }{2}-x \right)+{{\cos }^{4}}\left( 0+\dfrac{\pi }{2}-x \right)}dx}$
Now we know that $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x$ and $\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$
Hence we get
$\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( \dfrac{\pi }{2}-x \right)\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \\
 & =\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{\pi }{2}\cos x\sin x-x\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \\
\end{align}$
\[\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\dfrac{\pi }{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}\]
\[\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}=\dfrac{\pi }{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \\
 & \Rightarrow 2\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}=\dfrac{\pi }{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}=\dfrac{\pi }{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx} \\
\end{align}\]
Let us call \[\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}\] as I.
Hence now we have \[I=\dfrac{\pi }{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}............(1)\]
Now we know that
$\begin{align}
  & {{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}} \\
 & {{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab \\
\end{align}$
Hence using this we get
${{\sin }^{4}}x+{{\cos }^{4}}x={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x$
Substituting this in (1) we get.
$I=\dfrac{\pi }{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x\sin x}{(1)-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$
Now here we let us multiply the numerator by 2 and divide it by 2.
$I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos x\sin x}{1-2{{\cos }^{2}}x{{\sin }^{2}}x}dx}$
Also in the denominator we multiply by 2 and divide by 2 to the term $2{{\cos }^{2}}x{{\sin }^{2}}x$.
Hence now we get
$I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos x\sin x}{1-\dfrac{4{{\cos }^{2}}x{{\sin }^{2}}x}{2}}dx}$
$\Rightarrow I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos x\sin x}{1-\dfrac{{{(2\sin x\cos x)}^{2}}}{2}}dx}$
Now we know that $\sin 2x=2\sin x\cos x$. Using this formula we get
\[\Rightarrow I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{1-\dfrac{{{(\sin 2x)}^{2}}}{2}}dx}\]
Taking LCM in the denominator we get
\[\Rightarrow I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{\dfrac{2-({{\sin }^{2}}2x)}{2}}dx}\]
Now we know the trigonometric identity that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ .
\[\begin{align}
  & I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\sin 2x}{2-(1-{{\cos }^{2}}2x)}dx} \\
 & I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\sin 2x}{2-1+{{\cos }^{2}}2x}dx} \\
 & I=\dfrac{\pi }{8}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\sin 2x}{1+{{\cos }^{2}}2x}dx} \\
\end{align}\]
Now let us substitute $\cos 2x=t$ then we get $-2\sin 2x=dt$
Also note that as
$\begin{align}
  & x\to 0,t\to \cos 2(0)=1 \\
 & x\to \dfrac{\pi }{2}t\to \cos (2\dfrac{\pi }{2})=\cos \pi =-1 \\
\end{align}$
Hence using this in out integral we get
\[I=\dfrac{\pi }{8}\int\limits_{1}^{-1}{\dfrac{-dt}{1+{{t}^{2}}}dx}\]
We know that the integral $\int{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}x$.
\[\begin{align}
  & I=-\dfrac{\pi }{8}{{[{{\tan }^{-1}}x]}^{-1}}_{1} \\
 & I=-\dfrac{\pi }{8}\left[ {{\tan }^{-1}}(-1)-{{\tan }^{-1}}(1) \right] \\
 & I=-\dfrac{\pi }{8}\left[ -\dfrac{\pi }{4}-\dfrac{\pi }{4} \right] \\
 & I=-\dfrac{\pi }{8}\left[ \dfrac{-\pi }{2} \right] \\
 & I=\dfrac{{{\pi }^{2}}}{16} \\
\end{align}\]
Hence we have the value of given integral $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{x\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$ is $\dfrac{{{\pi }^{2}}}{16}$

Note: Now while using method of substitution for integration note that the limits of the integration also change. Hence if we substitute t as for f(x) check and the limits of x are given as a to b. then the new limit for t becomes f(a) to f(b).