Math Help - trigonometric integration

1. trigonometric integration

evaluate the following integrals, showing all working

a) int((sec^3*3x)(tan3x)) dx

b)int((sin15x)(cos9x)) dx

I was alright until the trig questions, could someone please get me started on these two.

2. Originally Posted by samdmansam
evaluate the following integrals, showing all working

a) int((sec^3*3x)(tan3x)) dx Mr F says: Make the substitution u = cos x.

b)int((sin15x)(cos9x)) dx Mr F says: 2 sin A cos B = sin (A + B) + sin (A - B) ......

I was alright until the trig questions, could someone please get me started on these two.
..

3. He4llo, samdmansam!

Another approach to (a) . . .

$a)\;\;\int \sec^3\!3x \tan3x\,dx$

$\text{Let: }\:u \:=\:\sec3x\quad\Rightarrow\quad du \:=\:3\sec3x\tan3x\,dx$

$\text{We have: }\;\int\underbrace{\sec^2\!3x}_{u^2}\underbrace{(\ sec3x\tan3x\,dx)}_{\frac{1}{3}du} \;=\;\frac{1}{3}\int u^2\,du$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

When we have: . $\int\sec^m\!x\tan^n\!x\,dx$
. . there is a procedure we can follow.

There are three cases to consider . . .

[1] Even secant, any tangent.

Example: . $\int\sec^4\!x\tan^3\!x\,dx$

$\text{"Peel off" }\sec^2\!x\text{ for our }du.$

We have: . $\int\sec^2\!x\,\tan^3\!x\,(\sec^2\!x\,dx)$
. . . . . . . . . . $\uparrow$
. . . . .
change to tangents
. . . . . . . . . $\downarrow$
. . $= \;\int(\tan^2\!x + 1)\tan^3\!x\,(\sec^2\!x\,dx) \;=\;\int(\tan^5\!x + \tan^3\!x)\,(\sec^2\!x\,dx)
$

$\text{Let: }\:u = \tan x\quad\Rightarrow\quad du = \sec^2\!x\,dx$

. . And we have: . $\int(u^5 + u^3)\,du$

[2] Odd tangent, any secant

Example: . $\int\sec^5\!x\tan^3\!x\,dx$

$\text{"Peel off" }\sec x\tan x\text{ for our }du.$

We have: . $\int\sec^4\!x\tan^2\!x\,(\sec x\tan x\,dx)$
. . . . . . . . . . . . . . $\uparrow$
. . . . . . . . .
change to secants
. . . . . . . . . . . . $\downarrow$
. . $= \;\int\sec^4\!x(\sec^2\!x-1)\,(\sec x\tan x\,dx) \;=\;\int(\sec^6\!x - \sec^4\!x)\,(\sec x\tan x\,dx)$

$\text{Let: }\,u \,=\,\sec x\quad\Rightarrow\quad du \,=\,\sec x\tan x\,dx$

. . And we have . $\int(u^6 - u^4)\,du$

[3] Odd secant, even tangent

This is the dreaded case, where a formula must be known
. . or it must be derived.

. . $\int\sec^3\!x\,dx \;=\;\frac{1}{2}\bigg[\sec x\tan x + \ln|\sec x + \tan x|\bigg] + C$

and a reduction formula is also recommended.