Question

Find integrate of (sin^2x-cos^2x)/(sinx×cosx) dx.​

Answer 1

We've to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{\sin^2x-\cos^2x}{\sin x\cos x}\ dx$

Multiply -1 inside and outside the integral and we get,

$\displaystyle\longrightarrow I=-\int\dfrac{\cos^2x-\sin^2x}{\sin x\cos x}\ dx$

Multiply and divide 2 on the integral as,

$\displaystyle\longrightarrow I=-2\int\dfrac{\cos^2x-\sin^2x}{2\sin x\cos x}\ dx$

We know that,

• $\cos^2x-\sin^2x=\cos(2x)$

• $2\sin x\cos x=\sin(2x)$

Thus,

$\displaystyle\longrightarrow I=-2\int\dfrac{\cos(2x)}{\sin(2x)}\ dx$

$\displaystyle\longrightarrow I=-\int\dfrac{\cos(2x)}{\sin(2x)}\cdot 2\ dx$

Let,

$\longrightarrow u=2x$

$\displaystyle\longrightarrow du=2\ dx$

Then the integral becomes,

$\displaystyle\longrightarrow I=-\int\dfrac{\cos u}{\sin u}\ du$

Since $\cos u$ is the derivative of $\sin u$ wrt $u,$

$\displaystyle\longrightarrow I=-\ln|\sin u|+C$

$\displaystyle\longrightarrow\underline{\underline{I=-\ln|\sin(2x)|+C}}$

Or,

$\displaystyle\longrightarrow\underline{\underline{I=\ln|\csc(2x)|+C}}$