# Question 5adb3

Sep 20, 2016

Alternative I :-

-{2cosx+ln|cscx-cotx|+C.

Alternative II :-

$\frac{1}{4} \left(2 x - \sin 2 x\right) - \ln | \csc x - \cot x | - \cos x + C$

#### Explanation:

Though the Question is not clear, but we solve it by taking $2$ possibilities :

Alternative I :-

int(sin^2x-cos^2x)/sinx dx=int{(sin^2x-(1-sin^2x)}/sinxdx

$= \int \frac{2 {\sin}^{2} x - 1}{\sin} x \mathrm{dx} = \int \left\{\frac{2 {\sin}^{2} x}{\sin} x - \frac{1}{\sin} x\right\} \mathrm{dx}$

$= 2 \int \sin x \mathrm{dx} - \int \csc x \mathrm{dx}$

$= 2 \left(- \cos x\right) - \ln | \csc x - \cot x |$

=-{2cosx+ln|cscx-cotx|+C.#

Alternative II :-

$\int \left\{{\sin}^{2} x - {\cos}^{2} \frac{x}{\sin} x\right\} \mathrm{dx}$

We use the Identity $: {\sin}^{2} x = \frac{1 - \cos 2 x}{2}$.

$\therefore \int \left\{{\sin}^{2} x - {\cos}^{2} \frac{x}{\sin} x\right\} \mathrm{dx}$

$= \int \frac{1 - \cos 2 x}{2} \mathrm{dx} - \int \frac{1 - {\sin}^{2} x}{\sin} x \mathrm{dx}$

$\frac{1}{2} \left(x - \sin \frac{2 x}{2}\right) - \int \left(\frac{1}{\sin} x - \sin x\right) \mathrm{dx}$

$= \frac{1}{4} \left(2 x - \sin 2 x\right) - \int \csc x \mathrm{dx} + \int \sin x \mathrm{dx}$

$= \frac{1}{4} \left(2 x - \sin 2 x\right) - \ln | \csc x - \cot x | - \cos x + C$

Enjoy Maths.!