# Use the indicated technique to evaluate the following integral. Show work detail to support your...

## Question:

Use the indicated technique to evaluate the following integral. Show work detail to support your solutions. Solving using other methods or with no detail is not acceptable.

{eq}\displaystyle\; \int \frac{x^{2}}{\sqrt{16 - x^{2}}}\,dx \quad\quad \left(\textrm{Trigonometric substitution}\right) {/eq}

## Indefinite Integral:

We can get the standard form of the given integrand by using the trigonometric substitution. Sine and Cosine are most used trigonometric function.

Then, we'll use the Integral reduction rule. The integration reduction rule states that {eq}\displaystyle \int \sin ^{n}x \ dx=-\dfrac {\sin ^{n-1}x \cos x}{n}+ \dfrac{n-1}{n} \int \sin^{n-2} x \ dx. {/eq} .

## Answer and Explanation:

We are given:

{eq}\displaystyle \int \frac{x^{2}}{\sqrt{16 - x^{2}}}\,dx {/eq}

Apply trigonometric substitution {eq}x= 4 \sin u \rightarrow \ dx = 4 \cos u \ d u {/eq}

{eq}= \displaystyle \int \frac{64\sin^2 u \cos u}{\sqrt{16-16 \sin^2 u }} \ du {/eq}

Apply the trigonometric identity {eq}\sin^2 u +\cos^2 u =1 {/eq}

{eq}= \displaystyle \int \frac{64\sin^2 u \cos u}{\sqrt{16 \cos^2 u }} \ du {/eq}

{eq}= \displaystyle \int \frac{64 \sin^2 u \cos u}{4 \cos u } \ du {/eq}

{eq}= \displaystyle \int 16 \sin^2 u \ du {/eq}

Apply the integral reduction rule, with {eq}n=2 {/eq}

{eq}\displaystyle= -16\dfrac {\cos u \sin u}{2}+ \dfrac{16}{2} \int 1 \ du {/eq}

{eq}\displaystyle= -4\sin 2u + 8u+C {/eq}

Substitute back {eq}x= 4 \sin u \Rightarrow u = \arcsin(x/4) {/eq}

{eq}\displaystyle= -4 sin( 2arcsin(x/4)) +8 \arcsin(x/4)+C {/eq}

Therefore, the solution is {eq}\displaystyle {\boxed{ \int \frac{x^{2}}{\sqrt{16 - x^{2}}}\,dx = -4 sin( 2arcsin(x/4)) +8 \arcsin(x/4)+C}} {/eq}

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from Calculus: Tutoring Solution

Chapter 7 / Lesson 14