# Question #36f17

Nov 25, 2017

$\int \frac{\sin 8 x \cdot \mathrm{dx}}{9 + {\left(\sin 4 x\right)}^{4}} = \frac{1}{12} \cdot \arctan \left({\left(\sin 4 x\right)}^{2} / 3\right) + C$

=$\frac{1}{12} \cdot \arctan \left(\frac{1 - \cos 8 x}{6}\right) + C$

#### Explanation:

$\int \frac{\sin 8 x \cdot \mathrm{dx}}{9 + {\left(\sin 4 x\right)}^{4}}$

=$\int \frac{2 \sin 4 x \cdot \cos 4 x \cdot \mathrm{dx}}{9 + {\left(\sin 4 x\right)}^{4}}$

=$\frac{1}{4} \cdot \int \frac{8 \sin 4 x \cdot \cos 4 x \cdot \mathrm{dx}}{9 + {\left(\sin 4 x\right)}^{4}}$

After using $u = {\left(\sin 4 x\right)}^{2}$ and $\mathrm{du} = 8 \sin 4 x \cdot \cos 4 x \cdot \mathrm{dx}$ transforms, this integral became,

$\frac{1}{4} \cdot \int \frac{\mathrm{du}}{9 + {u}^{2}}$

=$\frac{1}{12} \cdot \arctan \left(\frac{u}{3}\right) + C$

=$\frac{1}{12} \cdot \arctan \left({\left(\sin 4 x\right)}^{2} / 3\right) + C$

=$\frac{1}{12} \cdot \arctan \left(\frac{1 - \cos 8 x}{6}\right) + C$

Nov 25, 2017

$\int \setminus \sin \frac{8 x}{9 + {\sin}^{4} \left(4 x\right)} \setminus \mathrm{dx} = \frac{1}{12} {\tan}^{-} 1 \left(\frac{1}{3} {\sin}^{2} \left(4 x\right)\right) + C$

#### Explanation:

This integral is solved using loads of substitutions, so it is important that you keep track of what your substitutions are and what you are integrating with respect to.

Our first substitution will be $u = 4 x$ so we can reduce the sine argument to the top to $2 u$ and use the double angle identity.

The derivative of $u$ is $\frac{\mathrm{du}}{\mathrm{dx}} = 4$, so if we divide the integral by 4, we can integrate with respect to $u$:

$\int \setminus \sin \frac{8 x}{9 + {\sin}^{4} \left(4 x\right)} \setminus \mathrm{dx} = \frac{1}{4} \int \setminus \sin \frac{2 u}{9 + {\sin}^{4} \left(u\right)} \setminus \mathrm{du}$

Next, we will use the double angle identity, which says:
$\sin \left(2 \alpha\right) = 2 \sin \left(\alpha\right) \cos \left(\alpha\right)$

Applying this to the numerator of our integral, we get:
$\frac{1}{4} \int \setminus \frac{2 \sin \left(u\right) \cos \left(u\right)}{9 + {\sin}^{4} \left(u\right)} \setminus \mathrm{du} = \frac{2}{4} \int \setminus \frac{\sin \left(u\right) \cos \left(u\right)}{9 + {\sin}^{4} \left(u\right)} \setminus \mathrm{du}$

Now, we let $z = \sin \left(u\right)$, and to integrate with respect to $z$, we divide by the derivative, $\frac{\mathrm{dz}}{\mathrm{du}} = \cos \left(u\right)$:
$\frac{1}{2} \int \setminus \frac{z}{9 + {z}^{4}} \setminus \mathrm{dz}$

Time for another substitution. This time, we will let $t = {z}^{2}$. We divide through by $\frac{\mathrm{dt}}{\mathrm{dz}} = 2 z$ to get:
$\frac{1}{2} \cdot \frac{1}{2} \int \setminus \frac{1}{9 + {t}^{2}} \setminus \mathrm{dt}$

Now this is starting to resemble the integral of:
$\int \setminus \frac{1}{1 + {x}^{2}} \setminus \mathrm{dx} = {\tan}^{-} 1 \left(x\right) + C$
which we can easily solve. To get the integral we currently have into this form, we need to do one final substitution. This one will be so that $t = 3 r$. Notice that this time, we've defined our current variable in terms of an expression with another, so this time we will actually differentiate with respect to $r$, and multiply by that derivative.

$\frac{\mathrm{dt}}{\mathrm{dr}} = 3$, so let's multiply through:
$\frac{1}{4} \int \setminus \frac{3}{9 + {\left(3 r\right)}^{2}} \setminus \mathrm{dr} = \frac{3}{4} \int \setminus \frac{1}{9 + 9 {r}^{2}} \setminus \mathrm{dr}$

Now, we can factor out a $9$ in the denominator, which makes it so we can use our ${\tan}^{-} 1 \left(x\right)$ result from before:
$\frac{3}{4} \int \setminus \frac{1}{9 \left(1 + {r}^{2}\right)} \setminus \mathrm{dr} = \frac{3}{4} \cdot \frac{1}{9} \int \setminus \frac{1}{1 + {r}^{2}} \setminus \mathrm{dr} = \frac{3}{36} {\tan}^{-} 1 \left(r\right) + C$

And there we have it! All we need to do now is unwind all of the substitutions. Let's remember all of the variable definitions:
$u = 4 x$
$z = \sin \left(u\right)$
$t = {z}^{2}$
And the last one, $t = 3 r$, but here we actually need to solve for $r$, so we get:
$r = \frac{t}{3}$

Using these definitions, we get our final answer:
$\frac{1}{12} {\tan}^{-} 1 \left(\frac{1}{3} {\sin}^{2} \left(4 x\right)\right) + C$