Evaluate the integral using the given substitution.
-$$\int \csc ^ { 2 } 9 \theta \cot 9 \theta d \theta , \quad u = \cot 9 \theta$$
A) $- \frac { 1 } { 18 } \tan ^ { 2 } 9 \theta + C$
B) $- \frac { 1 } { 18 } \cot ^ { 2 } 9 \theta + C$
C) $\frac { 1 } { 18 } \cot ^ { 2 } \theta + C$
D) $\frac { 1 } { 6 } \csc ^ { 3 } 9 \theta \cot ^ { 2 } 9 \theta + C$

Question

Quizplus · Accepted Answer

Given $u = \cot 9\theta$, we differentiate both sides with respect to $\theta$ to find $du/d\theta$. The derivative of $\cot 9\theta$ is $-9\csc^2 9\theta$. Thus, $du = -9\csc^2 9\theta d\theta$. Rearranging, we get $d\theta = -\frac{1}{9} \csc^2 9\theta du$. Substituting $u$ and $d\theta$ into the integral gives $\int -\frac{1}{9} u du$, which integrates to $-\frac{1}{18}u^2 + C$. Substituting $u = \cot 9\theta$ back in gives $-\frac{1}{18}\cot^2 9\theta + C$.

Evaluate the Integral Using the Given Substitution ∫csc⁡29θcot⁡9θdθ,u=cot⁡9θ\int \csc ^ { 2 } 9 \theta \cot 9 \theta d \theta , \quad u = \cot 9 \theta∫csc29θcot9θdθ,u=cot9θ

Evaluate the Integral Using the Given Substitution $\int \csc ^ { 2 } 9 \theta \cot 9 \theta d \theta , \quad u = \cot 9 \theta$