What is \(\rm \int{{\left( \frac {1}{\cos^2 x} - \frac {1}{\sin^2 x} \right)}} dx\) equal to ?

Where c is the constant of integration

Question

Question

This question was previously asked in

Agniveer Vayu Group X 13 Oct 2023 Shift 1 Memory Based Paper

Answer 1

Concept:

\(\rm \int \sec^{2}xdx = \tan x + c \\ \rm \int cosec^{2}xdx = -\cot x+c\)

Calculation:

Let I = \(\rm \int{{\left( \frac {1}{\cos^2 x} - \frac {1}{\sin^2 x} \right)}} dx\)

\(= \rm \int \sec^{2}xdx - \rm \int cosec^{2}xdx\)

= tan x - (-cot x) + c

\(= \rm \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}+c \\ =\frac{\sin^2 x+ \cos^2 x}{\sin x \cos x}+c \\ = \frac{1}{\sin x \cos x}+c\)

\(= \rm \frac{2}{2\sin x \cos x} +c\\= \frac{2}{\sin2x}+c\) (∵ 2 sin x cos x = sin 2x)

= 2 cosec 2x + c