Consider the given integral I = $\int\limits_{0}^{\pi/2}$ 2 sin x cos x $\tan^{-1}$ (sin x) dx Let t = sinx ⇒ dt = cos x dx When x = $\frac{\pi}{2}$ , t = 1 When x = 0 , t = 0 Now, ∫ 2 sin x cos x $\tan^{-1}$ (sin x) dx = ∫ 2t $\tan^{-1}$ t dt = $[\tan^{-1} t]$ ∫ 2t dt - ∫ $\left[ \frac{d}{dt} (\tan^{-1} t) \int 2t \, dt \right]$ dt $[\tan^{-1} t]$ $\left[ 2 \cdot \frac{t^4}{2} \right]$ - ∫ $\left( \frac{1}{1+t^2} \times 2 \cdot \frac{t^2}{2} \right)$ dt = $t^2 \tan^{-1}$ t - ∫ $\frac{t^2}{1+t^2}$ dt = $t^2 \tan^{-1}$ t - ∫ $\left[ 1 - \frac{1}{1+t^2} \right]$ dt = $t^2 \tan^{-1}$ t - t + $$\tan^{-1} t \\ \therefore I =$∫↖{π/2}↙{0}$2 \sin x \cos x$tan^{-1}$(\sin x) \, dx \\ =$[t^2tan^{-1}t-t+tan^{-1}t]^1_0$\\ =$[1^2tan^{-1}1-1+tan^{-1}1]$-$[0^2tan^{-1}0-0+tan^{-1}0]$\\ =$[1× π/4 - 1 + π/4]$- 0 \\ =$π/4 - 1 + π/4$\\ =$π/2$- 1 \\ \textbf{OR} \\ I =$∫↖{π/2}↙{0}$ ${xsinxcosx}/{sin^4x+cos^4x}$dx \ldots (1) \\ \text{Using the property}$∫↖{a}↙{0}$f(x) \, dx =$∫↖{a}↙{0}$f(a-x) \, dx \\ I =$∫↖{π/2}↙{0}{(π/2-x)sin(π/2-x)cos(π/2-x)}/{sin^4 (π/2-x) + cos^4(π/2-x)}$dx \\ \Rightarrow I =$∫↖{π/2}↙{0}{(π/2 - x)cosxsinx}/{cos^4x+sin^4x}$dx \ldots (2) \\ \text{Adding (1) and (2)} \\ 2I =$∫↖{π/2}↙{0}(π/2 . sinxcosx)/{sin^4x+cos^4x}$dx \\ \Rightarrow I =$π/4 ∫↖{π/2}↙{0}|{sinxcosx}/{sin^4x+cos^4x}|$dx \\ =$π/4 ∫↖{π/2}↙{0}|{{sinxcosx}/{cos^4x}}/{{sin^4x}/{cos^4x}+1}|$dx \\$π/4 ∫↖{π/2}↙{0}{tanxsec^2x}/{tan^4x+1}$dx \\ \text{Put}$tan^2$x = z \\ \therefore 2 \tan x$sec^2$x \, dx = dz \\ \Rightarrow \tan x$sec^2$x \, dx =${dz}/2$\\ \text{When } x = 0,\; z = 0 \text{ and when } x =$π/2$,\; z = \infty \\ \therefore I =$π/4 ∫↖{∞}↙{0}{{dz}/2}/{z^2+1}$\\ \Rightarrow I =$π/8 ∫↖{∞}↙{0} {dz}/{1+z^2}$\\ =$π/8 |tan^{-1}(z)|^∞_0$\\ =$π/8 tan^{-1}∞-tan^{-1}0$\\ =$π/8(π/2-0)$\\ =$π^2/{16}$