1. ## integration by substitution

If someone could please help me solve some of these, because I'm getting stuck on all of them:

1. ∫ sin5xsin3x dx

2. ∫ (tanx + cot x) squared

3. ∫ (tan x) to the sixth power

4. ∫ (tan x)cubed times (sec x) cubed

thanks for any help!

2. Originally Posted by turtle
If someone could please help me solve some of these, because I'm getting stuck on all of them:

1. ∫ sin5xsin3x dx

2. ∫ (tanx + cot x) squared

3. ∫ (tan x) to the sixth power

4. ∫ (tan x)cubed times (sec x) cubed

thanks for any help!
My eyes are starting to get bleary, (it's been a long day), so I will just say the first one will go much easier if you note:
$sin(a)sin(b) = \frac{1}{2}(cos(a - b) - cos(a + b))$

-Dan

3. Originally Posted by turtle

3. ∫ (tan x) to the sixth power
$\int \tan^6 x dx$
Write,
$\int \tan^4 x \tan^2 x dx$
$\int \tan^4 x (\sec ^2 x-1) dx$
$\int \tan^4 x \sec ^2 x dx - \int \tan^4 x dx$
But,
$\int \tan^4 x dx= \int \tan^2 x(\sec^2 x -1) dx=\int \tan^2 x \sec^2 xdx - \int \tan^2 x dx$
But,
$\int \tan^2 xdx=\int \sec^2 x -1 dx$
Thus,
$\int \tan^4 x \sec^2 xdx-\int \tan^2 x\sec^2 dx+\int \sec^2 x dx-\int x dx$
You can do the first do using $u=\tan x$.
The third is simply $\tan x$.
The last is easy.
Thus,
$\frac{1}{5}\tan^5 x-\frac{1}{3}\tan^3 x-\tan x-\frac{1}{2}x^2+C$

4. Originally Posted by turtle

2. ∫ (tanx + cot x) squared
Open paranthesis,
$\int \tan^2 x+2\tan x\cot x+\cot^2 xdx$
Thus, (note that $\tan x\cot x=1$),

$\int \tan^2 x+1dx+\int cot^2 x +1dx$

$\int \sec^2 xdx +\int \csc^2 x dx$

$\tan x - \cot x +C$

5. Originally Posted by turtle

4. ∫ (tan x)cubed times (sec x) cubed
$\tan^3 x\sec^3 x=\tan^3 x\sec^2 x \sec x$
Call,
$u'=\tan^3 x\sec^2 x$
$v=\sec x$
Thus,
$u=\frac{1}{4}\tan^4 x$
$v'=-\sin x\cot^2 x$
The integral,
$\int \tan^3 x\sec^2 x \sec x dx$
Can be done by parts as above,
$\frac{1}{4}\sec x\tan^4 x +\frac{1}{4} \int \tan^4 x \sin x \cot ^2 x dx$
But,
$\tan^2 x\cot^2 x=1$
Thus,
$\frac{1}{4}\sec x\tan^4 x+\frac{1}{4} \int \tan^2 x\sin x dx$
The problem reduces to finding,

$\int \tan^2 x\sin x dx = \int \frac{\sin^3 x}{\cos^2 x} dx$
Write,
$\frac{\sin^3 x}{\cos^2 x}=\frac{(1-\cos^2 x)}{\cos ^2 x} \cdot \sin x$
Use,
$t=\cos x$
$t'=-\sin x$
Thus,
$\int -\frac{1-t^2}{t^2} t' dx$
$\int 1- t^{-2} dt$
$t+t^{-1}$
$\cos x+\sec x+C$

You can now do the full problem.