# How do you integrate int x^3/sqrt(64+x^2) by trigonometric substitution?

Jun 23, 2017

$I = \frac{64}{3} {\left({x}^{2} + 64\right)}^{\frac{3}{2}} - 64 \sqrt{{x}^{2} + 64} + C$

#### Explanation:

Use the substitution $x = 8 \tan \theta$. Then $\mathrm{dx} = 8 {\sec}^{2} \theta d \theta$.

$I = \int {\left(8 \tan \theta\right)}^{3} / \sqrt{64 + {\left(8 \tan \theta\right)}^{2}} \cdot 8 {\sec}^{2} \theta d \theta$

$I = \int \frac{512 {\tan}^{3} \theta}{\sqrt{64 + 64 {\tan}^{2} \theta}} \cdot 8 {\sec}^{2} \theta d \theta$

$I = \int \frac{512 {\tan}^{3} \theta}{\sqrt{64 \left(1 + {\tan}^{2} \theta\right)}} \cdot 8 {\sec}^{2} \theta d \theta$

Now use $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$.

$I = \int \frac{512 {\tan}^{3} \theta}{\sqrt{64 {\sec}^{2} \theta}} \cdot 8 {\sec}^{2} \theta d \theta$

$I = \int \frac{512 {\tan}^{3} \theta}{8 \sec \theta} \cdot 8 {\sec}^{2} \theta d \theta$

$I = \int 512 \sec \theta {\tan}^{3} \theta d \theta$

$I = \int 512 \sec \theta \tan \theta \left({\tan}^{2} \theta\right) d \theta$

We now use ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$.

$I = \int 512 \sec \theta \tan \theta \left({\sec}^{2} \theta - 1\right) d \theta$

Now let $u = \sec \theta$. Then $\mathrm{du} = \sec \theta \tan \theta d \theta$ and $d \theta = \frac{\mathrm{du}}{\sec \theta \tan \theta}$.

$I = \int 512 \sec \theta \tan \theta \cdot \left({u}^{2} - 1\right) \frac{\mathrm{du}}{\sec \theta \tan \theta}$

$I = \int \left(512 {u}^{2} - 512\right) \mathrm{du}$

$I = \frac{512}{3} {u}^{3} - 512 u + C$

$I = \frac{512}{3} {\sec}^{3} \theta - 512 \sec \theta$

From our initial substitution, we know that $\tan \theta = \frac{x}{8}$. Then $\sec \theta = \frac{\sqrt{{x}^{2} + 64}}{8}$

$I = \frac{64}{3} {\left({x}^{2} + 64\right)}^{\frac{3}{2}} - 64 \sqrt{{x}^{2} + 64} + C$

Hopefully this helps!

Jun 24, 2017

$\int \frac{{x}^{3}}{\sqrt{64 + {x}^{2}}} \text{d} x = \frac{1}{3} \sqrt{64 + {x}^{2}} \left({x}^{2} - 128\right) + C$.

#### Explanation:

I know you specified trigonometric substitution but I don't see why you'd use on in this case as a more obvious one stands out to me, because we have an ${x}^{2}$ with an ${x}^{3}$ on the numerator.

Someone's already submitted the trigonometric substitution method so thought I may as well share.

$I = \int \frac{{x}^{3}}{\sqrt{64 + {x}^{2}}} \text{d} x$.

Let $u = {x}^{2}$, $\left(\text{d"u)/("d} x\right) = 2 x$, $\text{d"x = 1/(2x) "d} u$.

Then,

$I = \frac{1}{2} \int {x}^{2} / \left(\sqrt{64 + {x}^{2}}\right) \text{d} u$,

$= \frac{1}{2} \int u {\left(64 + u\right)}^{- \frac{1}{2}} \text{d} u$.

At this point, I used integration by parts. Let $s = u$, and $\mathrm{dt} = {\left(64 + u\right)}^{- \frac{1}{2}} \mathrm{du}$. Then, for $\int s \mathrm{dt} = s t - \int t \mathrm{ds}$:

$I = \frac{1}{2} \left(\left[2 u {\left(64 + u\right)}^{\frac{1}{2}}\right] - 2 \int {\left(64 + u\right)}^{\frac{1}{2}} \text{d} u\right)$

$I = u {\left(64 + u\right)}^{\frac{1}{2}} - \frac{2}{3} {\left(64 + u\right)}^{\frac{3}{2}} + C$

Then, I factored out $\frac{1}{3} \sqrt{64 + {x}^{2}}$ and substituted $u = {x}^{2}$ back in:

$I = \frac{1}{3} \sqrt{64 + {x}^{2}} \left(3 {x}^{2} - 2 \left(64 + {x}^{2}\right)\right) + C$

$= \textcolor{b l u e}{\frac{1}{3} \sqrt{64 + {x}^{2}} \left({x}^{2} - 128\right) + C}$.

Jun 27, 2017

$\frac{1}{3} \sqrt{{x}^{2} + 64} \left({x}^{2} - 128\right) + C .$

#### Explanation:

Here is another way to solve the Problem, without using

substitution.

$I = \int {x}^{3} / \sqrt{{x}^{2} + 64} \mathrm{dx} ,$

$= \int \frac{{x}^{2} \cdot x}{\sqrt{{x}^{2} + 64}} \mathrm{dx} ,$

$= \int \frac{\left(\left({x}^{2} + 64\right) - 64\right) x}{\sqrt{{x}^{2} + 64}} \mathrm{dx} ,$

$= \int \left(\frac{{x}^{2} + 64}{\sqrt{{x}^{2} + 64}} - \frac{64}{\sqrt{{x}^{2} + 64}}\right) x \mathrm{dx} ,$

$= \int \left\{{\left({x}^{2} + 64\right)}^{\frac{1}{2}} - 64 {\left({x}^{2} + 64\right)}^{- \frac{1}{2}}\right\} \left\{\frac{1}{2} \frac{d}{\mathrm{dx}} \left({x}^{2} + 64\right)\right\} \mathrm{dx} ,$

$= \int {\left({x}^{2} + 64\right)}^{\frac{1}{2}} \left\{\frac{1}{2} \frac{d}{\mathrm{dx}} \left({x}^{2} + 64\right)\right\} \mathrm{dx}$
$- 64 \int {\left({x}^{2} + 64\right)}^{- \frac{1}{2}} \left\{\frac{1}{2} \frac{d}{\mathrm{dx}} \left({x}^{2} + 64\right)\right\} \mathrm{dx} ,$

$= \frac{1}{2} \int {\left({x}^{2} + 64\right)}^{\frac{1}{2}} \left(\frac{d}{\mathrm{dx}} \left({x}^{2} + 64\right)\right) \mathrm{dx}$
$- 32 \int {\left({x}^{2} + 64\right)}^{- \frac{1}{2}} \left(\frac{d}{\mathrm{dx}} \left({x}^{2} + 64\right)\right) \mathrm{dx} ,$

$= \frac{1}{2} {\left({x}^{2} + 64\right)}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) - 32 {\left({x}^{2} + 64\right)}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right) ,$

$= \frac{1}{3} {\left({x}^{2} + 64\right)}^{\frac{3}{2}} - 64 {\left({x}^{2} + 64\right)}^{\frac{1}{2}} ,$

=1/3sqrt(x^2+64){(x^2+64)-192)},

$= \frac{1}{3} \sqrt{{x}^{2} + 64} \left({x}^{2} - 128\right) + C .$

N.B.: The Final Integrals were obtained using the Rule,

$\int {\left[f \left(x\right)\right]}^{n} f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c , n \ne - 1.$

Enjoy Maths.!

Jul 9, 2017

After using $x = 8 \sinh u$ and $\mathrm{dx} = 8 \cosh u$ transforms, this integral became,

$\int {\left(8 \sinh u\right)}^{3} \cdot \frac{8 \cosh u}{8 \cosh u} \cdot \mathrm{du}$

$= \int 512 {\left(\sinh u\right)}^{3} \cdot \mathrm{du}$

$= \int 512 {\left(\sinh u\right)}^{2} \cdot \sinh u \cdot \mathrm{du}$

$= \int 512 \left({\left(\cosh u\right)}^{2} - 1\right) \cdot \sinh u \cdot \mathrm{du}$

$= \int 512 {\left(\cosh u\right)}^{2} \cdot \sinh u \cdot \mathrm{du} - \int 512 \sinh u \cdot \mathrm{du}$

$= \frac{512}{3} \cdot {\left(\cosh u\right)}^{3} - 512 \cosh u + C$

After using $\sinh u = \frac{x}{8}$ and $\cosh u = \frac{S q r t \left({x}^{2} + 64\right)}{8}$ inverse transforms, solution of this integral became,

=$\int \frac{{x}^{3}}{S q r t \left({x}^{2} + 64\right)} \cdot \mathrm{dx}$

=$= \frac{1}{3} \cdot {\left({x}^{2} + 64\right)}^{\frac{3}{2}} - 64 S q r t \left({x}^{2} + 64\right) + C$

=$= \frac{1}{3} \cdot \left({x}^{2} - 128\right) \cdot S q r t \left({x}^{2} + 64\right) + C$