Question

integral of [(sin²x - cos²x)/(sin²xcos²x)].dx is equal to
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) – tan x + cot x + C
(D) tanx + secx + C.

Answer 1

integral of [(sin^2x/{sin^2x cos^2x}-cos^2x/{sin^2x cos^2x})]

integral of [(1/cos^2x)-(1/sin^2x)]dx

infegral of [(sec^2x)-(cosec^2x)]dx

tanx-(-cotx)+c

tanx+cotx+c

Answer 2

HELLO DEAR,

GIVEN FUNCTION IS:- [(sin²x - cos²x)/(sin²xcos²)].dx

now, $\sf{\int{\frac{sin^2x }{sin^2xcos^2x} - \frac{cos^2x}{sin^2xcos^2x}}\,dx}$

$\sf{\Rightarrow \int{[ \frac{1}{cosx.cosx} - \frac{1}{sinx.sinx}]}\,dx}$

$\sf{\Rightarrow \int{(sec^2x - cosec^2x)}\,dx}$

$\sf{\Rightarrow tanx - (-cotx) + c}$

$\sf{\Rightarrow tanx +cotx + c}$


I HOPE ITS HELP YOU DEAR,
THANKS