# Thread: trig integration

1. ## trig integration

find each of the following integrals:

$\displaystyle \int tan^4 3x dx$

$\displaystyle \int sin^3 2x cos^5 2x dx$

$\displaystyle \int tan^3 x sec^5 x dx$

thanks : )

2. Originally Posted by qzno
find each of the following integrals:

$\displaystyle \int tan^4 3x dx$

$\displaystyle \int sin^3 2x cos^5 2x dx$

$\displaystyle \int tan^3 x sec^5 x dx$

thanks : )
$\displaystyle \tan^4(u) =$
$\displaystyle \tan^2(u) \cdot \tan^2(u) =$
$\displaystyle \tan^2(u)[\sec^2(u) - 1] =$
$\displaystyle \tan^2(u)\sec^2(u) - \tan^2(u) =$
$\displaystyle \tan^2(u)\sec^2(u) - \sec^2(u) + 1$

use substitution to integrate the first term, last two terms are straight-forward.

$\displaystyle \sin^3(u)\cos^5(u) =$
$\displaystyle \sin^3(u)[\cos^2(u)]^2 \cos(u) =$
$\displaystyle \sin^3(u)[1-\sin^2(u)]^2 \cos(u) =$
$\displaystyle [\sin^7(u) - 2\sin^5(u) + \sin^3(u)]\cos(u)$

straight substitution

$\displaystyle \tan^3{x}\sec^5{x} =$
$\displaystyle [\tan^2{x}\sec^4{x}]\sec{x}\tan{x} =$
$\displaystyle [(sec^2{x}-1)\sec^4{x}]\sec{x}\tan{x} =$
$\displaystyle [sec^6{x} - \sec^4{x}]\sec{x}\tan{x}$

straight substitution

3. thank you : )