Question

Integrate $\int \frac{{\sin }^3\left(\mathrm{x}\right)-{\cos }^3\left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$.

Answer 1

Step 1: Simplify the given integration.

In the question, $\int \frac{{\sin }^3\left(\mathrm{x}\right)-{\cos }^3\left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$ is given.

Let $I=\int \frac{{\sin }^3\left(\mathrm{x}\right)-{\cos }^3\left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$

$$\begin{gathered} I=\int \frac{{\sin }^3\left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}-\int \frac{{\cos }^3\left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx} \\ \Rightarrow I=\int \frac{\sin \left(\mathrm{x}\right)}{{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}-\int \frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\mathrm{dx} \end{gathered}$$

Assume that, $u=\int \frac{\sin \left(\mathrm{x}\right)}{{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx},v=\int \frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\mathrm{dx}$.

So, $I=u-v...\left(1\right)$

Step 2: Compute the value of $u$.

Since, $u=\int \frac{\sin \left(\mathrm{x}\right)}{{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$

Assume that, $\cos \left(\mathrm{x}\right)=\mathrm{t}$.

Differentiate both sides with respect to $x$.

$-\sin \left(\mathrm{x}\right)\mathrm{dx}=\mathrm{dt}$

So, $u$ can be written as $u=\int \frac{-\mathrm{dt}}{{\mathrm{t}}^2}$.

$$\begin{gathered} \mathrm{u}=\int \frac{-\mathrm{dt}}{{\mathrm{t}}^2}\left[\because \int {\mathrm{x}}^{-\mathrm{n}}\mathrm{dx}=\frac{{\mathrm{x}}^{1-\mathrm{n}}}{1-\mathrm{n}}\right] \\ \Rightarrow \mathrm{u}=\frac{1}{\mathrm{t}}+\mathrm{C} \\ \Rightarrow \mathrm{u}=\sec \left(\mathrm{x}\right)+\mathrm{C}\left[\because \cos \left(\mathrm{x}\right)=\frac{1}{\sec \left(\mathrm{x}\right)}\right] \end{gathered}$$

Step 3: Compute the value of $v$.

Since, $v=\int \frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\mathrm{dx}$

Assume that, $\sin \left(\mathrm{x}\right)=z$.

Differentiate both sides with respect to $x$.

$\mathrm{dz}=\cos \left(\mathrm{x}\right)\mathrm{dx}$

So, $v$ can be written as $\mathrm{v}=\int \frac{\mathrm{dz}}{{\mathrm{z}}^2}$.

$$\begin{gathered} \Rightarrow \mathrm{v}=\int \frac{\mathrm{dz}}{{\mathrm{z}}^2}\left[\because \int {\mathrm{x}}^{-\mathrm{n}}\mathrm{dx}=\frac{{\mathrm{x}}^{1-\mathrm{n}}}{1-\mathrm{n}}\right] \\ \Rightarrow \mathrm{v}=-\frac{1}{\mathrm{z}}+\mathrm{C} \\ \Rightarrow \mathrm{v}=-cosec\left(\mathrm{x}\right)+\mathrm{C}\left[\because \csc \left(\mathrm{x}\right)=\frac{1}{\sin \left(\mathrm{x}\right)}\right] \end{gathered}$$

Substitute the value of $u$ and $v$ in equation $1$.

$$\begin{gathered} \mathrm{I}=\mathrm{u}+\mathrm{v} \\ \Rightarrow \mathrm{I}=\sec \left(\mathrm{x}\right)+cosec\left(\mathrm{x}\right)+\mathrm{C} \end{gathered}$$

Hence $\int \frac{{\sin }^3\left(\mathrm{x}\right)-{\cos }^3\left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}=\sec \left(\mathrm{x}\right)+cosec\left(\mathrm{x}\right)+\mathrm{C}$.