Question

Evaluate the given indefinite integral: $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$?

Answer 1

Hint: We start solving the problem by converting the given integrand into subtraction of two different function and use the fact that $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$. We then use the facts that $\dfrac{1}{\cos x}=\sec x$, $\dfrac{1}{\sin x}=\operatorname{cosec}x$, \[\int{{{\sec }^{2}}xdx}=\tan x+C\] and $\int{{{\operatorname{cosec}}^{2}}xdx}=-\cot x+C$ to solve the problem further. We then use the facts $\cot x=\dfrac{1}{\tan x}$, $1+{{\tan }^{2}}x={{\sec }^{2}}x$, $\sec x=\dfrac{1}{\cos x}$, $\tan x=\dfrac{\sin x}{\cos x}$ and make necessary calculations to get the required answer.

Complete step-by-step solution
According to the problem, we need to find the solution for the given indefinite integral $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.
Let us assume the integral be I. So, we get $I=\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.
$\Rightarrow I=\int{\left( \dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}-\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \right)dx}$ ------(1).
We know that $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$, we use this result in equation (1).
$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}-\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.
$\Rightarrow I=\int{\dfrac{1}{{{\cos }^{2}}x}dx}-\int{\dfrac{1}{{{\sin }^{2}}x}dx}$ -----(2).
We know that $\dfrac{1}{\cos x}=\sec x$ and $\dfrac{1}{\sin x}=\operatorname{cosec}x$, We use these results in equation (2).
$\Rightarrow I=\int{{{\sec }^{2}}xdx}-\int{{{\operatorname{cosec}}^{2}}xdx}$ ------(3).
We know that \[\int{{{\sec }^{2}}xdx}=\tan x+C\] and $\int{{{\operatorname{cosec}}^{2}}xdx}=-\cot x+C$. We use these results in equation (3).
$\Rightarrow I=\tan x-\left( -\cot x \right)+C$.
$\Rightarrow I=\tan x+\cot x+C$ --------(4).
We know that $\cot x=\dfrac{1}{\tan x}$, we use this result in equation (4).
$\Rightarrow I=\tan x+\dfrac{1}{\tan x}+C$.
$\Rightarrow I=\dfrac{{{\tan }^{2}}x+1}{\tan x}+C$ ------(5).
We know that $1+{{\tan }^{2}}x={{\sec }^{2}}x$, we use this result in equation (5).
$\Rightarrow I=\dfrac{{{\sec }^{2}}x}{\tan x}+C$ -------(6).
We know that $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$. We use these results in equation (6).
$\Rightarrow I=\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{\sin x}{\cos x}}+C$.
$\Rightarrow I=\dfrac{1}{\sin x\cos x}+C$.
$\Rightarrow I=\dfrac{2}{2\sin x\cos x}+C$ -------(7).
We know that $\sin 2x=2\sin x\cos x$, we use this result in equation (7).
$\Rightarrow I=\dfrac{2}{\sin 2x}+C$.
So, we have found the result of indefinite integral $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$ as $\dfrac{2}{\sin 2x}+C$.
$\therefore$ $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\dfrac{2}{\sin 2x}+C$.

Note: We should not forget to add the constant of integration C while doing the problems related to Indefinite integrals. We can alternatively solve the problem as shown below:
$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.
$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\left( \sin x\cos x \right)}^{2}}}dx}$.
$\Rightarrow I=\int{\dfrac{-\cos 2x}{{{\left( \dfrac{\sin 2x}{2} \right)}^{2}}}dx}$.
$\Rightarrow I=\int{\dfrac{-4\cos 2x}{{{\sin }^{2}}2x}dx}$.
Let us assume $\sin 2x=t$, we get $dt=2\cos 2xdx$.
$\Rightarrow I=\int{\dfrac{-2}{{{t}^{2}}}dt}$.
$\Rightarrow I=\int{-2{{t}^{-2}}dt}$.
We know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$.
$\Rightarrow I=-2\times \left( \dfrac{{{t}^{-2+1}}}{-2+1} \right)+C$.
$\Rightarrow I=-2\times \left( \dfrac{{{t}^{-1}}}{-1} \right)+C$.
$\Rightarrow I=\dfrac{2}{t}+C$.
Now, let us substitute $t=\sin 2x$.
$\Rightarrow I=\dfrac{2}{\sin 2x}+C$.