# Integration by parts

• Nov 8th 2012, 11:09 AM
Greymalkin
Integration by parts
Attachment 25592How is it that the top becomes the bottom?, my textbook says the 1/2nd are simple(im lost), the third uses the identities $\displaystyle sin^2t=\frac{1-cos2t}{2}$ & $\displaystyle cos^2t=\frac{1+cos2t}{2}$, the last one uses the Pythagorean identity $\displaystyle sin^2t+cos^2t=1$, the last two seem doable what are the first two? I don't see how the 1/8 becomes 1/16 to start. Edit:ignore thumbnail, added new picture
• Nov 8th 2012, 01:43 PM
StefanTM
Re: Integration by parts
Hi Greymalkin,

when f(x)= cos(4x) => F(x)= 1/4* sin(4x) +C
g(x)= sin^2(2t)*cos2t => g(x)= sin(2t))^2*(sin(2t))' *1/2 => G(x)= 1/6*sin^3(2t) +C
Prove the results with derivation!
• Nov 9th 2012, 08:36 AM
Greymalkin
Re: Integration by parts
How does $\displaystyle -\frac{1}{8}\int cos^3 2tdt$ equal
$\displaystyle \frac{1}{8}sin^2 2tcos2tdt$$\displaystyle -\frac{1}{8}\int cos2t dt$ I get the last part, it is a factor, but how does the sin2 fit into there?
• Nov 10th 2012, 02:32 AM
StefanTM
Re: Integration by parts
Int=Integral
cos^2(2t)=1-sin^2(2t);
cos^3(2t)= cos^2(2t)*cos(2t)= (1-sin^2(2t))*cos(2t)= cos(2t) - sin^2(2t)*cos(2t)
True is:
Int cos^3(2t) dt = - Int sin^2(2t)*cos(2t)dt + Int cos(2t)dt =>
-1/8*Int cos^3(2t) dt =1/8* Int sin^2(2t)*cos(2t)dt -1/8* Int cos(2t)dt