# Few more integral questions.

• Jun 26th 2007, 07:31 PM
cinder
Few more integral questions.
First one, $\int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $u=x^5+2$ and $\frac{du}{5}=x^4dx$?

Second one, $\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
• Jun 26th 2007, 08:00 PM
ThePerfectHacker
Quote:

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

Yes substitute,
$u=x^5+2$ because its derivative appears almost outside.
Quote:

Second one, $\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
Hint: First substitute $t=\sqrt{\theta}$.

Hint: There is a second substitute :eek:
• Jun 26th 2007, 08:01 PM
Soroban
Hello, cinder!

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

Quote:

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

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

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

Quote:

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

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

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

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

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

• Jun 26th 2007, 08:11 PM
cinder
Quote:

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.
• Jun 26th 2007, 08:33 PM
CaptainBlack
Quote:

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

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

Observe that $5 x^4$ is the derivative of $x^5+2$ so:

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

For:

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

ask yourself what is the derivative of:

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

RonL
• Jun 26th 2007, 11:32 PM
red_dog
For the second integral you can make the substitution $\displaystyle \sin\sqrt{\theta}=u\Rightarrow \frac{\cos\sqrt{\theta}}{\sqrt{\theta}}d\theta=2du$.
Then $\displaystyle \int\frac{2}{u^2}du=-\frac{2}{u}+C$
So $\displaystyle \int\frac{\cos\sqrt{\theta}}{\sqrt{\theta}\sin ^2\sqrt{\theta}}d\theta=-\frac{2}{\sin\sqrt{\theta}}+C$