$\text { Find the indefinite integral by making the substitution } x = 9 \sin \theta \text {. }$
$\int \frac { 9 } { x ^ { 2 } \sqrt { 81 - x ^ { 2 } } } d x$
A) $- \frac { \sqrt { 81 - x ^ { 2 } } } { 81 x ^ { 2 } } + C$
B) $\frac { \sqrt { 81 - x ^ { 2 } } } { 81 x } + C$
C) $- \frac { \sqrt { 81 - x ^ { 2 } } } { 81 x } + C$
D) $-\frac{1}{81 x\left(81-x^{2}\right)^{3 / 2}}+C$
E)$\frac{1}{81 x\left(81-x^{2}\right)^{3 / 2}}+C$

Question

$\text { Find the indefinite integral by making the substitution } x = 9 \sin \theta \text {. }$ 
$\int \frac { 9 } { x ^ { 2 } \sqrt { 81 - x ^ { 2 } } } d x$
A) $- \frac { \sqrt { 81 - x ^ { 2 } } } { 81 x ^ { 2 } } + C$
B) $\frac { \sqrt { 81 - x ^ { 2 } } } { 81 x } + C$
C) $- \frac { \sqrt { 81 - x ^ { 2 } } } { 81 x } + C$
D) $-\frac{1}{81 x\left(81-x^{2}\right)^{3 / 2}}+C$ 
E)$\frac{1}{81 x\left(81-x^{2}\right)^{3 / 2}}+C$

Quizplus · Accepted Answer

Given $x = 9\sin\theta$, we have $dx = 9\cos\theta d\theta$. Substituting $x$ and $dx$ into the integral, we get $\int \frac{9}{(9\sin\theta)^2\sqrt{81-81\sin^2\theta}}9\cos\theta d\theta$. Simplifying, we use the identity $81-81\sin^2\theta = 81\cos^2\theta$, leading to $\int \frac{1}{81\sin^2\theta} d\theta$, which simplifies to $\int \csc^2\theta d\theta = -\cot\theta + C$. Substituting back for $x$, we get $-\frac{\sqrt{81-x^2}}{81x} + C$.

Find the indefinite integral by making the substitution x=9sin⁡θ. \text { Find the indefinite integral by making the substitution } x = 9 \sin \theta \text {. } Find the indefinite integral by making the substitution x=9sinθ.

$\text { Find the indefinite integral by making the substitution } x = 9 \sin \theta \text {. }$