# Integrate (16-1x^2)^0.51/8(16-1x^2)*-1 from 0 to 4

## \int_{0}^{4}\left(\sqrt{16-x^2}-\frac{1}{8}\cdot\left(16-x^2\right)\right)dx

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$16\int_{0}^{4}\sqrt{1-u^2}du-8+\frac{8}{3}$

## Step by step solution

Problem

$\int_{0}^{4}\left(\sqrt{16-x^2}-\frac{1}{8}\cdot\left(16-x^2\right)\right)dx$
1

Multiply $-1$ times $\frac{1}{8}$

$\int_{0}^{4}\left(\sqrt{16-x^2}-\frac{1}{8}\left(16-x^2\right)\right)dx$
2

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int_{0}^{4}\left(\frac{1}{8}x^2-2\right)dx+\int_{0}^{4}\sqrt{16-x^2}dx$
3

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int_{0}^{4}\sqrt{16-x^2}dx+\int_{0}^{4}\frac{1}{8}x^2dx+\int_{0}^{4}-2dx$
4

The integral of a constant is equal to the constant times the integral's variable

$\int_{0}^{4}\sqrt{16-x^2}dx+\int_{0}^{4}\frac{1}{8}x^2dx+\left[-2x\right]_{0}^{4}$
5

Evaluate the definite integral

$\int_{0}^{4}\sqrt{16-x^2}dx+\int_{0}^{4}\frac{1}{8}x^2dx-1\cdot 0\left(-2\right)+4\left(-2\right)$
6

Any expression multiplied by $0$ is equal to $0$

$\int_{0}^{4}\sqrt{16-x^2}dx+\int_{0}^{4}\frac{1}{8}x^2dx+0+4\left(-2\right)$
7

Multiply $-2$ times $4$

$\int_{0}^{4}\sqrt{16-x^2}dx+\int_{0}^{4}\frac{1}{8}x^2dx+0-8$
8

$x+0=x$, where $x$ is any expression

$\int_{0}^{4}\sqrt{16-x^2}dx+\int_{0}^{4}\frac{1}{8}x^2dx-8$
9

Taking the constant out of the integral

$\int_{0}^{4}\sqrt{16-x^2}dx+\frac{1}{8}\int_{0}^{4} x^2dx-8$
10

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\int_{0}^{4}\sqrt{16-x^2}dx+\left[\frac{1}{8}\cdot\frac{x^{3}}{3}\right]_{0}^{4}-8$
11

Simplify the fraction

$\int_{0}^{4}\sqrt{16-x^2}dx+\left[\frac{1}{24}x^{3}\right]_{0}^{4}-8$
12

Evaluate the definite integral

$\int_{0}^{4}\sqrt{16-x^2}dx-8-1\cdot 0^{3}\cdot \frac{1}{24}+4^{3}\cdot \frac{1}{24}$
13

Multiply $\frac{1}{24}$ times $-1$

$\int_{0}^{4}\sqrt{16-x^2}dx-8+0^{3}\left(-\frac{1}{24}\right)+4^{3}\cdot \frac{1}{24}$
14

Calculate the power

$\int_{0}^{4}\sqrt{16-x^2}dx-8+0\left(-\frac{1}{24}\right)+64\cdot \frac{1}{24}$
15

Any expression multiplied by $0$ is equal to $0$

$\int_{0}^{4}\sqrt{16-x^2}dx-8+0+64\cdot \frac{1}{24}$
16

Multiply $\frac{1}{24}$ times $64$

$\int_{0}^{4}\sqrt{16-x^2}dx-8+0+\frac{8}{3}$
17

Add the values $\frac{8}{3}$ and $0$

$\int_{0}^{4}\sqrt{16-x^2}dx-8+\frac{8}{3}$
18

Solve the integral $\int\sqrt{16-x^2}$ by trigonometric substitution using the substitution

$\begin{matrix}x=4\sin\left(\theta\right) \\ dx=4\cos\left(\theta\right)d\theta\end{matrix}$
19

Substituting in the original integral, we get

$\int_{0}^{4}4\sqrt{16-16\sin\left(\theta\right)^2}\cos\left(\theta\right)d\theta-8+\frac{8}{3}$
20

Taking the constant out of the integral

$4\int_{0}^{4}\sqrt{16-16\sin\left(\theta\right)^2}\cos\left(\theta\right)d\theta-8+\frac{8}{3}$
21

Solve the integral $\int\sqrt{16-16\sin\left(\theta\right)^2}\cos\left(\theta\right)d\theta$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=\sin\left(\theta\right) \\ du=\cos\left(\theta\right)d\theta\end{matrix}$
22

Isolate $d\theta$ in the previous equation

$\frac{du}{\cos\left(\theta\right)}=d\theta$
23

Substituting $u$ and $d\theta$ in the integral

$4\int_{0}^{4}\sqrt{16\left(1-u^2\right)}du-8+\frac{8}{3}$
24

The power of a product is equal to the product of it's factors raised to the same power

$4\int_{0}^{4}4\sqrt{1-u^2}du-8+\frac{8}{3}$
25

Taking the constant out of the integral

$4\cdot 4\int_{0}^{4}\sqrt{1-u^2}du-8+\frac{8}{3}$
26

Multiply $4$ times $4$

$16\int_{0}^{4}\sqrt{1-u^2}du-8+\frac{8}{3}$

$16\int_{0}^{4}\sqrt{1-u^2}du-8+\frac{8}{3}$

### Main topic:

Integration by trigonometric substitution

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