Kkutie7
  • Kkutie7
Check my work?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Kkutie7
  • Kkutie7
\[\int \frac{sin(4a)}{cos^{2}(4a)-cos(4a)}da\]
Kkutie7
  • Kkutie7
\[\rightarrow u=cos(4a) du=-4sin(4a)\rightarrow \frac{-1}{4}\int \frac{1}{u^{2}-u}du\]
Kkutie7
  • Kkutie7
\[\rightarrow \frac{-1}{4}\int \frac{1}{(u-(-1))((u-1))}du \rightarrow \] \[\frac{-1}{4}*\frac{-1}{2}(ln|u+1|-ln|u-1|)+c\rightarrow \] \[\frac{1}{8}(ln|cos(4a)+1|-ln|cos(4a)-1|)+c\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Kkutie7
  • Kkutie7
or \[\frac{1}{8}(ln|\frac{cos(4a)+1}{cos(4a)-1}|)+c\]
zepdrix
  • zepdrix
I don't quite understand this step:\[\rightarrow \frac{-1}{4}\int\limits \frac{1}{(u-(-1))((u-1))}du \rightarrow\]
zepdrix
  • zepdrix
What happened in the denominator there? Hmm
zepdrix
  • zepdrix
u^2-u should factor into u(1-u). Looks like you turned it into something else, I'm not quite sure 0_o
kanwal32
  • kanwal32
use partial fraction and take correct factors u(1-u)
Kkutie7
  • Kkutie7
i was trying to use my table so i did this: \[u^{2}-u=(u-1)(u+1)\]
kanwal32
  • kanwal32
\[\frac{ 1 }{ u-1 }-\frac{ 1 }{ u }\]
kanwal32
  • kanwal32
u^2-u taking common u u(u-1) easy way
Kkutie7
  • Kkutie7
ok well \[\rightarrow u=cos(4a) du=-4sin(4a)\rightarrow \frac{-1}{4}\int \frac{1}{u^{2}-u}du\rightarrow \frac{-1}{4}\int \frac{1}{u(u-1)}du\] \[\frac{1}{u(u-1)}=\frac{A}{u}+\frac{B}{u-1}\] \[1=A(u-1)+Bu\rightarrow x=0, 1=-A, A=-1\rightarrow x=1, 1=B\] \[\frac{-1}{4} (\int \frac{-1}{u}du+\int\frac{1}{u-1}du)\] \[\frac{-1}{4}(ln|u|+ln|u-1|)+C\] \[\frac{-1}{4}(ln|cos(4a)-1|-ln|cos(4a)|)+C\]
Kkutie7
  • Kkutie7
The answer is supposed to be \[\frac{1}{4}ln(cos(4a)-2ln(sin(4a)))+C\]
kanwal32
  • kanwal32
integration of -1/u=-lnu
Kkutie7
  • Kkutie7
I know
kanwal32
  • kanwal32
is it sin2a or sin4a
Kkutie7
  • Kkutie7
4a where did I write 2a?
Kkutie7
  • Kkutie7
i'm confused as to why I didn't get the correct answer

Looking for something else?

Not the answer you are looking for? Search for more explanations.