Question

The value of $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}} \,dx$ is
A. $\dfrac{{{\pi ^2}}}{4}$
B. $\dfrac{{{\pi ^2}}}{8}$
C. $\dfrac{{{\pi ^2}}}{{32}}$
D. $\dfrac{{3{\pi ^2}}}{{16}}$

Answer 1

Hint : Here the question is related to the integration. The integral is of the form of definite integral where the limit points are mentioned. The function for the integration is trigonometric function to integrate the function we solve by the substitution method. Hence we obtain the required solution for the given question.

Complete step by step solution:
Now consider the given integral as I
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}} \,dx$ ---------- (1)
Now replace the x by $\left( {\dfrac{\pi }{2} - x} \right)$
So it can be written as
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\dfrac{\pi }{2} - x} \right)\sin \left( {\dfrac{\pi }{2} - x} \right)\cos \left( {\dfrac{\pi }{2} - x} \right)}}{{{{\sin }^4}\left( {\dfrac{\pi }{2} - x} \right) + {{\cos }^4}\left( {\dfrac{\pi }{2} - x} \right)}}} \,dx$
As we know that $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$ and $\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$
Therefore the above equation can be written as
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\dfrac{\pi }{2} - x} \right)\cos x\sin x}}{{{{\cos }^4}x + {{\sin }^4}x}}} \,dx$ ------ (2)
On adding the equation (1) and (2) we get $\Rightarrow 2I = \dfrac{\pi }{2}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x\sin x}}{{{{\cos }^4}x + {{\sin }^4}x}}} \,dx$
Put ${\sin ^2}x = t$ , on differentiating this function we get $2\sin x\cos xdx = dt$ . When the function changes the limit points will also change
When x = 0 then t = 0 and when $x = \dfrac{\pi }{2}$ then t = 1
Therefore the above equation can be written as
$\Rightarrow 2I = \dfrac{\pi }{2} \times \dfrac{1}{2}\int\limits_0^1 {\dfrac{{dt}}{{{{(1 - t)}^2} + {t^2}}}} \,$
On simplifying we get
$\Rightarrow 2I = \dfrac{\pi }{4}\int\limits_0^1 {\dfrac{{dt}}{{1 + {t^2} - 2t + {t^2}}}} \,$
Take 2 to RHS we get
$\Rightarrow I = \dfrac{\pi }{8}\int\limits_0^1 {\dfrac{{dt}}{{1 + 2{t^2} - 2t}}} \,$
Simplify the denominator
$\Rightarrow I = \dfrac{\pi }{{16}}\int\limits_0^1 {\dfrac{{dt}}{{\dfrac{1}{2} + {t^2} - t}}} \,$
The denominator term can be replaced by ${\left( {t - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4}$ , so the equation or integral can be written as
$\Rightarrow I = \dfrac{\pi }{{16}}\int\limits_0^1 {\dfrac{{dt}}{{{{\left( {t - \dfrac{1}{2}} \right)}^2} + \dfrac{1}{4}}}} \,$
On applying integration we have
$\left. { \Rightarrow I = \dfrac{\pi }{{16}} \times \dfrac{1}{{\dfrac{1}{2}}}{{\tan }^{ - 1}}\left( {\dfrac{{t - \dfrac{1}{2}}}{{\dfrac{1}{2}}}} \right)} \right|_0^1$
Apply the limit points we get
$\Rightarrow I = \dfrac{\pi }{{16}} \times \dfrac{1}{{\dfrac{1}{2}}}{\tan ^{ - 1}}\left( {\dfrac{{(1 - 0) - \dfrac{1}{2}}}{{\dfrac{1}{2}}}} \right)$
On simplifying we get
$\Rightarrow I = \dfrac{\pi }{8}{\tan ^{ - 1}}\left( {\dfrac{{1 - \dfrac{1}{2}}}{{\dfrac{1}{2}}}} \right)$
$\Rightarrow I = \dfrac{\pi }{8}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}}}} \right)$
Since the numerator and the denominator is same we cancel them
$\Rightarrow I = \dfrac{\pi }{8}{\tan ^{ - 1}}\left( 1 \right)$
The value of ${\tan ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{4}$ , therefore we get
$\Rightarrow I = \dfrac{\pi }{8} \times \dfrac{\pi }{4}$
On simplification
$\Rightarrow I = \dfrac{{{\pi ^2}}}{{32}}$
Hence we have obtained the solution
The option C is the correct one
So, the correct answer is “Option C”.

Note : By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.