Question

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

• A. $2\text{cosec}2x+c$ where $c$ is the constant of integration
• B. $-2\cot 2x+c$ where $c$ is the constant of integration
• C. $2\sec 2x+c$ where $c$ is the constant of integration
• D. $-2\tan 2x+c$ where $c$ is the constant of integration

Answer 1

Answer: (A)

$2\text{cosec}2x+c$ where $c$ is the constant of integration

$\int (\frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x})dx$

$=\int \left(\frac{{\sin }^{2}x-{\cos }^{2}x}{{\cos }^{2}x{\sin }^{2}x}\right)dx$

$=\int \frac{-\cos 2x}{\frac{4{\cos }^{2}x{\sin }^{2}x}{4}}dx$

$=\int \left(\frac{-4\cos 2x}{{\sin }^{2}2x}\right)dx$

$=\int -4\cot 2x\text{cosec}2xdx$

$=2\text{cosec}2x+c$