Integrate the function.
-$$\int \sqrt { 16 - x ^ { 2 } } d x$$
A) $8 \sin ^ { - 1 } \left( \frac { x } { 4 } \right) + \frac { x \sqrt { 16 - x ^ { 2 } } } { 2 } + C$
B) $\frac { x } { 16 \sqrt { 16 - x ^ { 2 } } } + \frac { \sqrt { 16 - x ^ { 2 } } } { x } + C$
C) $\frac { 16 x } { \sqrt { 16 - x ^ { 2 } } } + \frac { x } { 2 } + C$
D) $8 x - \frac { x \sqrt { 16 - x ^ { 2 } } } { 2 } + C$

Question

Quizplus · Accepted Answer

Use the substitution $x=4\sin	heta$, then $dx=4\cos	heta d	heta$ and $\sqrt{16-x^2}=4\cos	heta$. Substituting these in the integral, we have:
\begin{align*} \int \sqrt{16-x^2}dx &= \int 4\cos^2	heta\cdot 4\cos	heta d	heta \ &=\int 4(1-\sin^2	heta)\cdot 4\cos	heta d	heta \ &=\int 16\cos	heta d	heta-\int 16\sin^2	heta\cdot\cos	heta d	heta \ &= 16\sin	heta -\int 16(1-\cos^2	heta)\cdot\cos	heta d	heta\ &=16\sin	heta-\int(16\cos	heta-16\cos^3	heta)d	heta\ &=16\sin	heta-16\sin	heta+4\sin^3	heta\cos	heta \ &=8\sin^{-1}\left(\frac{x}{4}ight) + \frac{x\sqrt{16-x^2}}{2} + C. \end{align*}

Integrate the Function ∫16−x2dx\int \sqrt { 16 - x ^ { 2 } } d x∫16−x2​dx

Integrate the Function $\int \sqrt { 16 - x ^ { 2 } } d x$