Question

Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .

Answer 1

Let I = \[\int\frac{dx}{\sin^2 x \cos^2 x}\]

Dividing the numerator and denominator by cos⁴ x, we get:

I = \[\int\frac{se c^2 x \cdot se c^2 x}{\tan^2 x}dx\]

\[\int\frac{\left( 1 + \tan^2 x \right) \cdot se c^2 x}{\tan^2 x}dx\]

Put tan x = t

⇒ \[se c^2 xdx = dt\]

∴ I = \[\int\frac{1 + t^2}{t^2}dt\] = \[\int1dt + \int\frac{1}{t^2}dt\]

⇒ I = t −\[\frac{1}{t}\] + C

⇒ I = tan x − cot x + C

∴ \[\int\frac{dx}{\sin^2 x \cos^2 x}\] = tan x − cot x + C