anonymous
  • anonymous
Integral sin(2x) (cos(x))^2 dx Substitution rule o by parts ?
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
sin(2x)=2sin(x)cos(x) use this
myininaya
  • myininaya
\[2 \int\limits_{}^{}\sin(x)(\cos(x))^3 dx\]
myininaya
  • myininaya
use sub u=cos(x)

Looking for something else?

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

More answers

myininaya
  • myininaya
let me know if you need more help
anonymous
  • anonymous
thx a lot
myininaya
  • myininaya
great! you got the rest? :)
anonymous
  • anonymous
yeah i guess
myininaya
  • myininaya
ok well just let me know if you run into troubles
anonymous
  • anonymous
k
anonymous
  • anonymous
the solution is -1/2cos^4 (x) +c
myininaya
  • myininaya
\[u=\cos(x)\] => \[ du=-\sin(x) dx\] \[-2 \int\limits_{}^{} u^3 du=-2 \cdot \frac{u^4}{4}+C=-\frac{1}{2} u^4+C=-\frac{1}{2}\cos^4(x)+C\]
myininaya
  • myininaya
yes thats right! :)
anonymous
  • anonymous
thx
anonymous
  • anonymous
gor the help
anonymous
  • anonymous
for+*

Looking for something else?

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