Question

Solve $\int_{0}^{\pi/2}\dfrac{xsin~xcos~x}{cos^{4}~x+sin^{4}x}.\text{d}x $ [IIT 1985]
(A) 0
(B)$ \dfrac{\pi}{8}$
(C) $\dfrac{\pi^{2}}{8}$
(D) $\dfrac{\pi^{2}}{16}$

Answer 1

Hint: The process of combining smaller parts into a single, integrated system is known as integration. It is the inverse of differentiation, so it is also known as anti differentiation.



Formula Used:The fundamental property of integration $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$is used to solve this question. It is a question of definite integration, so the upper and lower limits are substituted after solving the integration.



Complete step by step solution:The given integral is: I = $\int_{0}^{\pi/2}\dfrac{xsin~xcos~x}{cos^{4}~x+sin^{4}x}.\text{d}x $ ------(1)
Here the given function is f(x) = $\dfrac{xsin~xcos~x}{cos^{4}~x+sin^{4}x}$ and the total range over which the integral is to be done is 0 to $\pi/2$, with a lower limit of 0 and an upper limit $\pi/2$.
We know that the fundamental property of integrals is $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$
Using this property, we get
I = $\int_{0}^{\dfrac{\pi}{2}}\dfrac{(\dfrac{\pi}{2}-x)sin~(\dfrac{\pi}{2}-x)cos~(\dfrac{\pi}{2}-x)}{cos^{4}~(\dfrac{\pi}{2}-x)+sin^{4}(\dfrac{\pi}{2}-x)}.\text{d}x$
We know that $\\sin(\dfrac{\pi}{2}-x)=cos~x$ and $\\cos(\dfrac{\pi}{2}-x)=sin~x$.
I = $\int_{0}^{\dfrac{\pi}{2}}\dfrac{(\dfrac{\pi}{2}-x)cos~xsin~x}{sin^{4}~x+cos^{4}x}.\text{d}x$ ----- (2)
Adding equations (1) and (2), we get
2I = $\dfrac{\pi}{2}\int_{0}^{\pi/2}\dfrac{cos~xsin~x}{sin^{4}~x+cos^{4}x}.\text{d}x$
If we substitute $\tan~x= \dfrac{sin~x}{cos~x}$ and $\dfrac{1}{\cos^{2}~x}=sec^{2}~x$, we get
2I = $\dfrac{\pi}{2}\int_{0}^{\dfrac{\pi}{2}}\dfrac{tan~xsec^{2}~x}{1+tan^{4}~x}.\text{d}x$
Putting $\\tan^{2}~x=t$
Differentiating both sides, $\\2tan~x.sec^{2}~x=\text{d}t$
At x = 0, t = 0
At x = $\dfrac{\pi}{2}$, t = $\infty$
2I = $\dfrac{\pi}{4}\int_{0}^{\infty}\dfrac{1}{1+t^{2}}.\text{d}t$
2I = $\dfrac{\pi}{4}\left[\tan^{-1}~t\right]_{0}^{infty}$
Putting values of $\tan^{-1}~t$ at t = 0 and t = $infty$ .
2I = $\dfrac{\pi}{4}\left[\dfrac{\pi}{2}-0\right]$
2I = $\dfrac{\pi^{2}}{8}$
I = $\dfrac{\pi^{2}}{16}$
Hence, $\int_{0}^{\pi/2}\dfrac{xsin~xcos~x}{cos^{4}~x+sin^{4}x}.\text{d}x$ over the range 0 to $\dfrac{\pi}{2}$ is $\dfrac{\pi^{2}}{16}$.



Option ‘D’ is correct


Note: It should be remembered that after putting the values of upper and lower limit is also changed as the function is changed. The integration should be done carefully. The values of the trigonometric should be substituted carefully.