∫ dx/sin^2x cos^2x = (A) tan x + cot x + C (B) tan2 x − cot x + C

Question

\(\int \frac{dx}{sin^2x\ cos^2 x} =\)

(A) tan x + cot x + C

(B) tan2 x − cot x + C

(C) − cot x + tan x + C

(D) tan x ∙ cot x + C

Answer 1

Correct option is (C) -cot x + tan x + C

\(\int \frac{dx}{\sin^2x.\cos^2x}\)

\(= \int \frac{(\sin^2x + \cos^2x)}{\sin^2x.\cos^2x}\)

\(= \int \frac{\sin^2x}{\sin^2x. \cos^2x} dx + \int \frac{\cos^2 x}{\sin^2x. \cos^2x}dx\)

\(=\int\frac{dx}{\cos^2x}dx+\int \frac{dx}{\sin^2x}\)

\(= \int \sec^2xdx+ \int \text{cosec}^2x dx\)

\(= \tan x - \cot x+ C\)

\(= - \cot x+\tan x+ C\)

Answer 2

Correct option is (C) − cot x + tan x + C