Mathematics - Indefinite Integration Question with Solution | TestHub

Let $I=\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\left({\sin }^{3}x+{\cos }^{3}x\right)}^2}dx$ $$\begin{array}{rl}& \mathrm{I}=\int {\left(\frac{\sin x\cdot \cos x}{{\sin }^{3}x+{\cos }^{3}x}\right)}^{2}dx\\ & \mathrm{I}=\int {\left(\frac{\sin x\cdot \cos x}{{\cos }^{3}x\left(1+{\tan }^{3}x\right)}\right)}^{2}dx\\ & =\int {\left(\frac{\sin x\cdot {\sec }^{2}x}{\left(1+{\tan }^{3}x\right)}\right)}^{2}dx\\ & \text{ Put }1+{\tan }^{3}x=\mathrm{t}\\ & dt=3{\tan }^{2}x{\sec }^{2}xdx\text{ or }dx\\ & =\frac{dt}{3{\tan }^{2}x{\sec }^{2}x}\\ & \therefore \mathrm{I}=\int \frac{{\sin }^{2}x\cdot {\sec }^{4}x}{t^2}\times \frac{dt}{3{\tan }^{2}x{\sec }^{2}x}\end{array}$$ $$\begin{array}{rl}& \mathrm{I}=\frac{1}{3}\int \frac{{\sin }^{2}x\cdot {\sec }^{4}x}{t^2}\times \frac{dt}{\frac{{\sin }^{2}x}{{\cos }^{2}x}\times {\sec }^{2}x}\\ & =\frac{1}{3}\int \frac{{\sin }^{2}x\cdot {\sec }^{4}x}{t^2}\times \frac{dt}{{\sin }^{2}x{\sec }^{4}x}\\ & \therefore \mathrm{I}=\frac{1}{3}\int \frac{dt}{t^2}=\frac{1}{3}\int t^{-2}dt\\ & \mathrm{I}=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c=\frac{-1}{3}\left[\frac{1}{t}\right]+c\\ & \text{ or }\mathrm{I}=-\frac{1}{3\left(1+{\tan }^{3}x\right)}+c\end{array}$$