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!
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!
$\displaystyle \int \tan^6 x dx$
Write,
$\displaystyle \int \tan^4 x \tan^2 x dx$
$\displaystyle \int \tan^4 x (\sec ^2 x-1) dx$
$\displaystyle \int \tan^4 x \sec ^2 x dx - \int \tan^4 x dx$
But,
$\displaystyle \int \tan^4 x dx= \int \tan^2 x(\sec^2 x -1) dx=\int \tan^2 x \sec^2 xdx - \int \tan^2 x dx$
But,
$\displaystyle \int \tan^2 xdx=\int \sec^2 x -1 dx$
Thus,
$\displaystyle \int \tan^4 x \sec^2 xdx-\int \tan^2 x\sec^2 dx+\int \sec^2 x dx-\int x dx$
You can do the first do using $\displaystyle u=\tan x$.
The third is simply $\displaystyle \tan x$.
The last is easy.
Thus,
$\displaystyle \frac{1}{5}\tan^5 x-\frac{1}{3}\tan^3 x-\tan x-\frac{1}{2}x^2+C$
$\displaystyle \tan^3 x\sec^3 x=\tan^3 x\sec^2 x \sec x$
Call,
$\displaystyle u'=\tan^3 x\sec^2 x$
$\displaystyle v=\sec x$
Thus,
$\displaystyle u=\frac{1}{4}\tan^4 x$
$\displaystyle v'=-\sin x\cot^2 x$
The integral,
$\displaystyle \int \tan^3 x\sec^2 x \sec x dx$
Can be done by parts as above,
$\displaystyle \frac{1}{4}\sec x\tan^4 x +\frac{1}{4} \int \tan^4 x \sin x \cot ^2 x dx$
But,
$\displaystyle \tan^2 x\cot^2 x=1$
Thus,
$\displaystyle \frac{1}{4}\sec x\tan^4 x+\frac{1}{4} \int \tan^2 x\sin x dx$
The problem reduces to finding,
$\displaystyle \int \tan^2 x\sin x dx = \int \frac{\sin^3 x}{\cos^2 x} dx$
Write,
$\displaystyle \frac{\sin^3 x}{\cos^2 x}=\frac{(1-\cos^2 x)}{\cos ^2 x} \cdot \sin x$
Use,
$\displaystyle t=\cos x$
$\displaystyle t'=-\sin x$
Thus,
$\displaystyle \int -\frac{1-t^2}{t^2} t' dx$
$\displaystyle \int 1- t^{-2} dt$
$\displaystyle t+t^{-1}$
$\displaystyle \cos x+\sec x+C$
You can now do the full problem.