Question

$\int (\tan x-\cot x{)}^{2}dx=$

• A. $\tan x+x+C$
• B. $\tan x-x+C$
• C. $\tan x-\cot x+C$
• D. $\tan x-\cot x-4x+C$

Answer 1

The correct option is D $\tan x-\cot x-4x+C$

$I=\int {(\tan x-\cot x)}^{2}dx=\int \frac{{({\tan }^{2}x-1)}^2}{{\tan }^{2}x}dx=\int \frac{{\tan }^{4}x+1-2{\tan }^{2}x}{{\tan }^{2}x}dx$

$=\int {\tan }^{2}xdx+\int {\cot }^{2}xdx-2\int dx=\int [({\tan }^{2}x+1)-1]dx+\int \frac{{\cos }^{2}x}{{\sin }^{2}x}dx-2x+C$

$=\int {\sec }^{2}xdx-\int dx+\int \frac{1-{\sin }^{2}x}{{\sin }^{2}x}dx-2x+C$

$=\tan x-x+\int co{\sec }^{2}xdx-\int dx-2x+C=\tan x-4x+\int co{\sec }^{2}xdx+C=\tan x-4x-\cot x+C$

$I=\tan x-\cot x-4x+C$