# Thread: Few more integral questions.

1. ## Few more integral questions.

First one, $\displaystyle \int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $\displaystyle x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $\displaystyle u=x^5+2$ and $\displaystyle \frac{du}{5}=x^4dx$?

Second one, $\displaystyle \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.

2. Originally Posted by cinder
First one, $\displaystyle \int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $\displaystyle x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $\displaystyle u=x^5+2$ and $\displaystyle \frac{du}{5}=x^4dx$?
Yes substitute,
$\displaystyle u=x^5+2$ because its derivative appears almost outside.
Second one, $\displaystyle \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
Hint: First substitute $\displaystyle t=\sqrt{\theta}$.

Hint: There is a second substitute

3. Hello, cinder!

Looks like you're a bit shaky about substitution . . .

$\displaystyle \int x^4(x^5+2)^7\,dx$
Let $\displaystyle u = x^5+ 2\quad\Rightarrow\quad du = 5x^4\,dx\quad\Rightarrow\quad dx = \frac{du}{5x^4}$

Substitute: .$\displaystyle \int x^4\cdot u^7\left(\frac{du}{5x^4}\right) \;=\;\frac{1}{5}\int u^7\,du \;= \;\frac{1}{40}u^8 + C$

Back-substitute: .$\displaystyle \frac{1}{40}(x^5 + 2)^8 + C$

$\displaystyle \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2\!(\ sqrt{\theta})} d\theta$

Let $\displaystyle \sqrt{\theta} = u\quad\Rightarrow\quad \theta = u^2\quad\Rightarrow\quad d\theta = 2u\,du$

Substitute: .$\displaystyle \int\frac{\cos u}{u\sin^2\!u}(2u\,du) \;=\;2\int\frac{\cos u}{\sin^2\!u}\,du \;=\;2\int\left(\frac{1}{\sin u}\!\cdot\!\frac{\cos u}{\sin u}\right)\,du$

. . $\displaystyle = \;2\int\csc u\cot u\,du \;=\;-2\csc u + C$

Back-substitute: .$\displaystyle -2\csc(\sqrt{\theta}) + C$

4. Originally Posted by Soroban
Hello, cinder!

Looks like you're a bit shaky about substitution . . .
Yes, I'm just not sure what I should be substituting.

5. Originally Posted by cinder
First one, $\displaystyle \int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $\displaystyle x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $\displaystyle u=x^5+2$ and $\displaystyle \frac{du}{5}=x^4dx$?

Second one, $\displaystyle \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
Observe that $\displaystyle 5 x^4$ is the derivative of $\displaystyle x^5+2$ so:

$\displaystyle \int x^4(x^5+2)^7dx=\frac{1}{5}\int \left[\frac{d}{dx}(x^5+2)\right](x^5+2)^7dx=\frac{1}{40}(x^5+2)^8$

For:

$\displaystyle \int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$,

ask yourself what is the derivative of:

$\displaystyle \frac{1}{\sin(\sqrt{\theta})}$

RonL

6. For the second integral you can make the substitution $\displaystyle \displaystyle \sin\sqrt{\theta}=u\Rightarrow \frac{\cos\sqrt{\theta}}{\sqrt{\theta}}d\theta=2du$.
Then $\displaystyle \displaystyle \int\frac{2}{u^2}du=-\frac{2}{u}+C$
So $\displaystyle \displaystyle \int\frac{\cos\sqrt{\theta}}{\sqrt{\theta}\sin ^2\sqrt{\theta}}d\theta=-\frac{2}{\sin\sqrt{\theta}}+C$