JEE Main 2014 — Mathematics Calculus

Let $\displaystyle I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$I=∫(sin3x+cos3x)2sin2xcos2x​dx $$\begin{aligned} &\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\sin ^3 x+\cos ^3 x}\right)^2 d x \\ &\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\cos ^3 x\left(1+\tan ^3 x\right)}\right)^2 d x \\ &=\int\left(\frac{\sin x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)}\right)^2 d x \\ &\text { Put } 1+\tan ^3 x=\mathrm{t} \\ &d t=3 \tan ^2 x \sec ^2 x d x \text { or } d x \\ &=\frac{d t}{3 \tan ^2 x \sec ^2 x} \\ &\therefore \mathrm{I}=\int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{3 \tan ^2 x \sec ^2 x} \end{aligned}$$​I=∫(sin3x+cos3xsinx⋅cosx​)2dxI=∫(cos3x(1+tan3x)sinx⋅cosx​)2dx=∫((1+tan3x)sinx⋅sec2x​)2dx Put 1+tan3x=tdt=3tan2xsec2xdx or dx=3tan2xsec2xdt​∴I=∫t2sin2x⋅sec4x​×3tan2xsec2xdt​​ $$\begin{aligned} &\mathrm{I}=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\frac{\sin ^2 x}{\cos ^2 x} \times \sec ^2 x} \\ &=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\sin ^2 x \sec ^4 x} \\ &\therefore \mathrm{I}=\frac{1}{3} \int \frac{d t}{t^2}=\frac{1}{3} \int t^{-2} d t \\ &\mathrm{I}=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c=\frac{-1}{3}\left[\frac{1}{t}\right]+c \\ &\text { or } \mathrm{I}=-\frac{1}{3\left(1+\tan ^3 x\right)}+c \end{aligned}$$​I=31​∫t2sin2x⋅sec4x​×cos2xsin2x​×sec2xdt​=31​∫t2sin2x⋅sec4x​×sin2xsec4xdt​∴I=31​∫t2dt​=31​∫t−2dtI=31​[−2+1t−2+1​]+c=3−1​[t1​]+c or I=−3(1+tan3x)1​+c​