# Integrate -1(16+x^2)^0.5*11 from -41*1 to 1

## \int_{\left(-41\cdot1\right)}^{1}11\left(-1\right)\sqrt{16+x^2}dx

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$-176\int_{-41}^{1}\sqrt{u^2+1}du$

## Step by step solution

Problem

$\int_{\left(-41\cdot1\right)}^{1}11\left(-1\right)\sqrt{16+x^2}dx$
1

Multiply $11$ times $-1$

$\int_{-41}^{1}-11\sqrt{x^2+16}dx$
2

Taking the constant out of the integral

$-11\int_{-41}^{1}\sqrt{x^2+16}dx$
3

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

$\begin{matrix}x=4\tan\left(\theta\right) \\ dx=4\sec\left(\theta\right)^2d\theta\end{matrix}$
4

Substituting in the original integral, we get

$-11\int_{-41}^{1}4\sqrt{16\tan\left(\theta\right)^2+16}\sec\left(\theta\right)^2d\theta$
5

Taking the constant out of the integral

$-11\cdot 4\int_{-41}^{1}\sqrt{16\tan\left(\theta\right)^2+16}\sec\left(\theta\right)^2d\theta$
6

Multiply $4$ times $-11$

$-44\int_{-41}^{1}\sqrt{16\tan\left(\theta\right)^2+16}\sec\left(\theta\right)^2d\theta$
7

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

$\begin{matrix}u=\tan\left(\theta\right) \\ du=\sec\left(\theta\right)^2d\theta\end{matrix}$
8

Isolate $d\theta$ in the previous equation

$\frac{du}{\sec\left(\theta\right)^2}=d\theta$
9

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

$-44\int_{-41}^{1}\sqrt{16\left(u^2+1\right)}du$
10

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

$-44\int_{-41}^{1}4\sqrt{u^2+1}du$
11

Taking the constant out of the integral

$-44\cdot 4\int_{-41}^{1}\sqrt{u^2+1}du$
12

Multiply $4$ times $-44$

$-176\int_{-41}^{1}\sqrt{u^2+1}du$

$-176\int_{-41}^{1}\sqrt{u^2+1}du$

### Main topic:

Integration by trigonometric substitution

0.44 seconds

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