1. ## Basic Integration

I've been studying for a integration and volumes test. I can do most of them but these few are giving me headaches. You don't have to solve them for me, just tell me what I need to do. Any help is appreciated.

1)
$\int \sqrt{16+x^2} \,dx$

I know I need to substitute $x = 4\,tan\,\theta \Rightarrow dx = 4\,sec^2\theta \,d\theta$
$\int 4sec\theta \bullet 4sec^2\theta \,d\theta$
= $\int 16 sec^3\theta\,d\theta$

I don't know how to go on from here.
The answer is $\frac{1}{2}x\sqrt{16+x^2} + 8 ln (x+\sqrt{16+x^2})$

2)
$\int \frac{1}{x^2\sqrt{x-1}}dx$

I've tried letting $u= \sqrt{x-1}$, but got nowhere from that.

3)
$\int_0^{\frac{\pi}{2}} \sqrt{1+sin\,2x}\, dx$

no clue

2. Originally Posted by Gusbob
2) $\int \frac{1}{x^2\sqrt{x-1}}dx$

I've tried letting $u= \sqrt{x-1}$, but got nowhere from that.
Try $x = sec^2 \theta$

Originally Posted by Gusbob
3) $\int_0^{\frac{\pi}{2}} \sqrt{1+sin\,2x}\, dx$

no clue
Did you observe $\sqrt{1+sin\,2x} = |\cos x + \sin x|$ ?

3. 1.
Let $u=\sin(x)$

$\int \frac{dx}{\cos^3(x)}=\int \frac{\cos(x)}{\cos^4(x)}dx=\int \frac{du}{(1-u^2)^2}$

$\frac{1}{(1-u^2)}=\frac{1}{2}\cdot{\left(\frac{1}{1-u}+\frac{1}{1+u}\right)}$ (1)

square (1) and apply (1) again with what you get

2.
After that sub you get $2\int\frac{du}{(u^2+1)^2}$

Let: $z=\arctan(u)$ and remember that $\frac{1}{\tan^2(z)+1}=\cos^2(z)$

4. Did you observe $\sqrt{1+sin\,2x} = |\cos x + \sin x|$ ?
How does that work?

5. Hello,

Originally Posted by Gusbob
How does that work?
$1=\cos^2x+\sin^2x$

$\sin 2x=2 \cos x \sin x$

$\implies 1+\sin 2x=\cos^2x+\sin^2x+2 \cos x \sin x=(\cos x+\sin x)^2$

6. Originally Posted by Gusbob
I've been studying for a integration and volumes test. I can do most of them but these few are giving me headaches. You don't have to solve them for me, just tell me what I need to do. Any help is appreciated.

1)
$\int \sqrt{16+x^2} \,dx$

I know I need to substitute $x = 4\,tan\,\theta \Rightarrow dx = 4\,sec^2\theta \,d\theta$
$\int 4sec\theta \bullet 4sec^2\theta \,d\theta$
= $\int 16 sec^3\theta\,d\theta$

I don't know how to go on from here.
The answer is $\frac{1}{2}x\sqrt{16+x^2} + 8 ln (x+\sqrt{16+x^2})$

2)
$\int \frac{1}{x^2\sqrt{x-1}}dx$

I've tried letting $u= \sqrt{x-1}$, but got nowhere from that.

3)
$\int_0^{\frac{\pi}{2}} \sqrt{1+sin\,2x}\, dx$

no clue
For the first one by trig sub we get

Let $x=4\tan(\theta)\Rightarrow{\theta=arctan\bigg(\fra c{x}{4}\bigg)}$

This means that $dx=16\sec^2(\theta)$

Imputting this we get

$\int\sqrt{16+16\tan^2(\theta)}\sec^2(\theta)d\thet a=4\int\sec^3(\theta)d\theta$

Now to integrate that thing

Ok so we do this

remember that $\sec^3(x)=(1+\tan^2(x))\sec(x)=\sec(x)+\sec(x)\tan ^2(x)$

So subbing this in remember waht this is actually equal to we get

$\int\sec(\theta)+\sec(\theta)\tan^2(\theta)d\theta =\ln|\sec(\theta)+\tan(\theta)|+\int\sec(\theta)\t an^2(\theta)d\theta$

Now evaluating the integral within their we let

$dv=\sec(\theta)\tan(\theta)d\theta$

and let $u=\tan(\theta)$

So $v=sec(\theta)$

and $du=\sec^2(\theta)d\theta$

Applying parts we get

$\int\sec(\theta)\tan^2(\theta)d\theta=\sec(\theta) \tan(\theta)-\int\sec^3(\theta)d\theta$

Putting this back into the integral we get

$\int\sec(\theta)+\sec(\theta)\tan^2(\theta)d\theta =\int\sec^3(\theta)d\theta=$ $\ln|\sec(\theta)+\tan(\theta)|+\sec(\theta)\tan(\t heta)-\int\sec^3(\theta)d\theta$ $\Rightarrow{2\int\sec^3(\theta)d\theta=\ln|\sec(\t heta)+\tan(\theta)|+\sec(\theta)\tan(\theta)}$

Dividing both sides by two we get

$16\int\sec^3(\theta)d\theta=2\bigg[\ln|\sec(\theta)+\tan(\theta)|+\sec(\theta)\tan(\t heta)\bigg]+C$

Now remembering that $\theta=artan\bigg(\frac{x}{4}\bigg)$

we need to imput back in...so seeing that
$\sec\bigg(arctan\bigg(\frac{x}{4}\bigg)\bigg)=\fra c{\sqrt{x^2+16}}{4}$

and that $\tan\bigg(arctan\bigg(\frac{x}{4}\bigg)\bigg)=\fra c{x}{4}$

wee see that

$\int\sqrt{x^2+16}dx=16\bigg[\frac{1}{2}\bigg[\ln\bigg|\frac{\sqrt{x^2+16}}{4}+\frac{x}{4}\bigg| +\frac{\sqrt{x^2+16}}{4}\cdot\frac{x}{4}\bigg]+C\bigg]=$ $8\bigg[\ln\bigg|\frac{\sqrt{x^2+16}}{4}+\frac{x}{4}\bigg| +\frac{x\sqrt{x^2+16}}{16}\bigg]+C\bigg]$