# How do you integrate int x^3 sqrt(-x^2 - 8x-41)dx using trigonometric substitution?

Aug 7, 2018

intx^3sqrt(-x^2-8x-41)dx=5^5i*(1/5sec(theta)^5−1/3sec(theta)^3)-5^4*3/2i(2sec(theta)^3tan(theta)-sec(theta)tan(theta)-ln(|sec(theta)+tan(theta)|))+16isec(theta)^3-5^2*32i(sec(theta)tan(theta)+ln(|sec(theta)+tan(theta)|)+C,

$C \in \mathbb{R}$,

$\theta = {\tan}^{- 1} \left(\frac{x + 4}{5}\right)$,

$\sec \left({\tan}^{- 1} \left(u\right)\right) = \sqrt{{u}^{2} + 1}$, here $u = \frac{x + 4}{5}$

See explanations below.

#### Explanation:

$I = \int {x}^{3} \sqrt{- {x}^{2} - 8 x - 41} \mathrm{dx}$

$= i \int {x}^{3} \sqrt{{x}^{2} + 8 x + 41} \mathrm{dx}$

Complete the square:

$I = i \int {x}^{3} \sqrt{{\left(x + 4\right)}^{2} + 25} \mathrm{dx}$

$= i \int {x}^{3} \sqrt{{\left(x + 4\right)}^{2} + {5}^{2}} \mathrm{dx}$

Now let $x + 4 = 5 \tan \left(\theta\right)$

$\mathrm{dx} = 5 \sec {\left(\theta\right)}^{2} d \theta$

So:

$I = i \int {\left(5 \tan \left(\theta\right) - 4\right)}^{3} \sqrt{{5}^{2} \left(\tan {\left(\theta\right)}^{2} + 1\right)} \cdot 5 \sec {\left(\theta\right)}^{2} d \theta$

Because $\tan {\left(\theta\right)}^{2} + 1 = \sec {\left(\theta\right)}^{2}$

$= 5 i \int {\left(5 \tan \left(\theta\right) - 4\right)}^{3} \cdot 5 \sec {\left(\theta\right)}^{3} d \theta$

$= 25 i \int {\left(5 \tan \left(\theta\right) - 4\right)}^{3} \sec {\left(\theta\right)}^{3} d \theta$

Now, ${\left(a - b\right)}^{3} = {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3}$, and here, $a = 5 \tan \left(\theta\right)$, $b = 4$

So:

$I = {5}^{2} i \int \left({5}^{3} \tan {\left(\theta\right)}^{3} - {5}^{2} \cdot 12 \tan {\left(\theta\right)}^{2} + 5 \cdot 48 \tan \left(\theta\right) - 64\right) \sec {\left(\theta\right)}^{3} d \theta$

$= {5}^{5} i$$\int \tan {\left(\theta\right)}^{3} \sec {\left(\theta\right)}^{3} d \theta$$- {5}^{4} \cdot 12 i$$\int \tan {\left(\theta\right)}^{2} \sec {\left(\theta\right)}^{3} d \theta$$+ {5}^{3} \cdot 48 i$$\int \tan \left(\theta\right) \sec {\left(\theta\right)}^{3} d \theta$$- {5}^{2} \cdot 64 i$$\int \sec {\left(\theta\right)}^{3} d \theta$

Well, at this point, you have a sum of defined integral, you just have to sum
( click on each integral to see explanations associated ).

I=5^5i*(1/5sec(theta)^5−1/3sec(theta)^3)-5^4*3/2i(2sec(theta)^3tan(theta)-sec(theta)tan(theta)-ln(|sec(theta)+tan(theta)|))+16isec(theta)^3-5^2*32i(sec(theta)tan(theta)+ln(|sec(theta)+tan(theta)|)+C, $C \in \mathbb{R}$

Finally, $\theta = {\tan}^{- 1} \left(\frac{x + 4}{5}\right)$, and $\sec \left({\tan}^{- 1} \left(u\right)\right) = \sqrt{{u}^{2} + 1}$, here $u = \frac{x + 4}{5}$

I let you find what is the writing of $I$ using $x$ and not $\theta$, it's very long but easy.