Question

Trigonometric Substitution

Am I correct on this one ?
https://www.dropbox.com/s/tevq8fzrxc8qd5w/Screenshot%202013-10-27%2002.30.02.jpg

Answer 1

I think you should have \[8 \int\limits_{}^{}(1-\cos(2 \u) du \text{ and not } 2 \int\limits_{}^{}(1-\cos(2 u) d \u\]
Also \[\int\limits_{}^{}\cos(2 u ) du=\frac{1}{2} \sin(2
u) +C \text{ and not } \int\limits_{}^{}\cos(2u)d u= \sin(2u) +C\]

Answer 2

hm

Answer 3

I understood the second line

Answer 4

but not the first one

Answer 5

like why it should be 8 outside of the integral

Answer 6

from where

Answer 7

and how the u is in the denominator of the angle ?

Answer 8

It isn't...That is a type-o. I made a error in the latex coding.

Answer 9

Like I was using theta at first and I decided to change it to u since you used u.

Answer 10

I need /theta to make the \[\theta \] show

Answer 11

\[\int\limits_{}^{}\frac{ 16 \sin^2(u) }{ \sqrt{16 \cos^2(u)}} \cdot 4 \cos(u) d u=\int\limits_{}^{}\frac{16\sin^2(u)}{4 \cos(u)} 4 \cos(u) du\]

Answer 12

Do you see the 4cos(u)/4cos(u)=1?

Answer 13

\[\sin^2(u)=\frac{1}{2}(1-\cos(2 u)) => 16 \sin^2(u)=16 \cdot \frac{1}{2}(1-\cos(2u))=8(1-\cos(2u))\]