Question

Check my work?

Answer 1

\[\int \frac{sin(4a)}{cos^{2}(4a)-cos(4a)}da\]

Answer 2

\[\rightarrow u=cos(4a) du=-4sin(4a)\rightarrow \frac{-1}{4}\int \frac{1}{u^{2}-u}du\]

Answer 3

\[\rightarrow \frac{-1}{4}\int \frac{1}{(u-(-1))((u-1))}du \rightarrow \]
\[\frac{-1}{4}*\frac{-1}{2}(ln|u+1|-ln|u-1|)+c\rightarrow \]
\[\frac{1}{8}(ln|cos(4a)+1|-ln|cos(4a)-1|)+c\]

Answer 4

or \[\frac{1}{8}(ln|\frac{cos(4a)+1}{cos(4a)-1}|)+c\]

Answer 5

I don't quite understand this step:\[\rightarrow \frac{-1}{4}\int\limits \frac{1}{(u-(-1))((u-1))}du \rightarrow\]

Answer 6

What happened in the denominator there? Hmm

Answer 7

u^2-u should factor into u(1-u).
Looks like you turned it into something else, I'm not quite sure 0_o

Answer 8

use partial fraction and take correct factors u(1-u)

Answer 9

i was trying to use my table so i did this:
\[u^{2}-u=(u-1)(u+1)\]

Answer 10

\[\frac{ 1 }{ u-1 }-\frac{ 1 }{ u }\]

Answer 11

u^2-u taking common u
u(u-1) easy way

Answer 12

ok well
\[\rightarrow u=cos(4a) du=-4sin(4a)\rightarrow \frac{-1}{4}\int \frac{1}{u^{2}-u}du\rightarrow \frac{-1}{4}\int \frac{1}{u(u-1)}du\]
\[\frac{1}{u(u-1)}=\frac{A}{u}+\frac{B}{u-1}\]
\[1=A(u-1)+Bu\rightarrow x=0, 1=-A, A=-1\rightarrow x=1, 1=B\]
\[\frac{-1}{4} (\int \frac{-1}{u}du+\int\frac{1}{u-1}du)\]
\[\frac{-1}{4}(ln|u|+ln|u-1|)+C\]
\[\frac{-1}{4}(ln|cos(4a)-1|-ln|cos(4a)|)+C\]

Answer 13

The answer is supposed to be \[\frac{1}{4}ln(cos(4a)-2ln(sin(4a)))+C\]

Answer 14

integration of -1/u=-lnu

Answer 15

I know

Answer 16

is it sin2a or sin4a

Answer 17

4a where did I write 2a?

Answer 18

i'm confused as to why I didn't get the correct answer