# Integrate (36-4x^2)^0.5

## \int\sqrt{36-4x^2}dx

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$18\int\sqrt{1-u^2}du$

## Step by step solution

Problem

$\int\sqrt{36-4x^2}dx$
1

First, factor the terms inside the radical for an easier handling

$\int\sqrt{4\left(9-x^2\right)}dx$
2

Taking the constant out of the radical

$\int2\sqrt{9-x^2}dx$
3

Solve the integral $\int2\sqrt{9-x^2}$ by trigonometric substitution using the substitution

$\begin{matrix}x=3\sin\left(\theta\right) \\ dx=3\cos\left(\theta\right)d\theta\end{matrix}$
4

Substituting in the original integral, we get

$\int6\cos\left(\theta\right)\sqrt{9-9\sin\left(\theta\right)^2}d\theta$
5

Taking the constant out of the integral

$6\int\cos\left(\theta\right)\sqrt{9-9\sin\left(\theta\right)^2}d\theta$
6

Solve the integral $\int\frac{u\sqrt{9\left(1-\sin\left(\theta\right)^2\right)}}{-\sin\left(\theta\right)}du$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=\sin\left(\theta\right) \\ du=\cos\left(\theta\right)d\theta\end{matrix}$
7

Isolate $d\theta$ in the previous equation

$\frac{du}{\cos\left(\theta\right)}=d\theta$
8

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

$6\int\sqrt{9\left(1-u^2\right)}du$
9

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

$6\int3\sqrt{1-u^2}du$
10

Taking the constant out of the integral

$6\cdot 3\int\sqrt{1-u^2}du$
11

Multiply $3$ times $6$

$18\int\sqrt{1-u^2}du$

$18\int\sqrt{1-u^2}du$

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### Main topic:

Integration by trigonometric substitution

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