Thread: [SOLVED] ∫ xcos^2(2x)dx & ∫ln(x^2+14x+49)dx

Hi,
I would like some help evaluating the following integrals.
I'm not sure if I am doing all the right steps.


∫$\displaystyle xcos^{2}(2x)dx$

∫$\displaystyle ln(x^2+14x+49)dx$

Thank you

Originally Posted by kinto

Hi,
I would like some help evaluating the following integrals.
I'm not sure if I am doing all the right steps.


∫$\displaystyle xcos^{2}(2x)dx$

∫$\displaystyle ln(x^2+14x+49)dx$

Thank you

Show us the steps you've done and where you think you may be wrong.
The first one may be easily done by integration by parts knowing that INT(cos^2(t) dt) = [x + sin x cos x]/2.
The second one requires from you to know INT(ln x)dx , which can be done straight integrating by parts, and also to put x^2 + 14x + 49 = (x+7)^2 and use some basic properties of logaritms.

Tonio

For the first one, I tried replacing my $\displaystyle cos^2(2x)$ by using a half-angle identity

∫$\displaystyle x((1+cos4x)/2)dx$
1/2∫$\displaystyle x(1+cos4x)$

u=x
du=dx

dv=cos4x
v=$\displaystyle -1/4sin4x$

$\displaystyle 1/2[(x^2/2)+x((-1/4)sin4x)-(1/4)cos4x]$

Originally Posted by kinto

Hi,
I would like some help evaluating the following integrals.
I'm not sure if I am doing all the right steps.


∫$\displaystyle xcos^{2}(2x)dx$

∫$\displaystyle ln(x^2+14x+49)dx$

Thank you

For the second one, try simplifying x^2+14x+49. You might see the rest of the steps after .