The antiderivative $\int \left( 1 + \cot ^ { 2 } x \right) d x$ is

A) $x + \frac { \cot ^ { 3 } x } { 3 } + C$
B) $x - \frac { \cot ^ { 3 } x } { 3 } + C$
C) $- \csc ^ { 2 } x + C$
D) $- \cot ^ { 2 } x + C$
E) $- \cot x + C$

Question

The antiderivative $\int \left( 1 + \cot ^ { 2 } x \right) d x$ is

A) $x + \frac { \cot ^ { 3 } x } { 3 } + C$
B) $x - \frac { \cot ^ { 3 } x } { 3 } + C$
C) $- \csc ^ { 2 } x + C$
D) $- \cot ^ { 2 } x + C$
E) $- \cot x + C$

Quizplus · Accepted Answer

The antiderivative $\int \left( 1 + \cot ^ { 2 } x \right) d x$ simplifies to $\int \csc^2 x dx$ using the trigonometric identity $1 + \cot^2 x = \csc^2 x$. The antiderivative of $\csc^2 x$ is $-\cot x + C$, making option E correct.

The Antiderivative ∫(1+cot⁡2x)dx\int \left( 1 + \cot ^ { 2 } x \right) d x∫(1+cot2x)dx

The Antiderivative $\int \left( 1 + \cot ^ { 2 } x \right) d x$