• Jul 29th 2009, 09:21 PM
qzack96
I really need some help with these trig substitution problems. Please write the u and du values for each. I'd really really appreciate it !

1) Integrate [e^x/(e^2x +1)] dx

2) Integrate [X^5 (lnx)] dx

3) Integrate [sinx*cosx] dx

Thank you Again !!
• Jul 29th 2009, 10:02 PM
pickslides
Quote:

Originally Posted by qzack96

3) Integrate [sinx*cosx] dx

$\displaystyle \int \sin(x)\cos(x) dx$

$\displaystyle u = \sin(x) \Rightarrow \frac{du}{dx} = \cos(x)$

$\displaystyle \int u \frac{du}{dx}dx$

$\displaystyle \int u~du$

can you take it from here?
• Jul 29th 2009, 10:05 PM
matheagle
1 let $\displaystyle u=e^x$, so your integral is now

$\displaystyle \int {du\over u^2+1}=\arctan u+C=\arctan (e^x)+C$

2 Parts with $\displaystyle u=\ln x$ and $\displaystyle dv=x^5$

3 can be done several ways, $\displaystyle u=\sin x$, $\displaystyle w=\cos x$ or use $\displaystyle \sin (2x)=2\sin x\cos x$

The last way makes this

$\displaystyle {1\over 2}\int \sin (2x)dx={-1\over 4} \cos (2x)+C$
• Jul 30th 2009, 01:37 AM
enjam
For 1, we can divide both sides by e^x to get:
∫ 1 / (e^x + e^-x) . dX
Have you been taught that cosh(x) = 0.5 (e^x + e^-x)?
So we have:
∫ 1 / 2cosh(x) . dx
= 0.5 ∫ 1 / cosh(x) . dx
= 0.5 ∫ sech(x) . dx
Are you able to go on from here or have I just made it even more complicated?
• Jul 30th 2009, 03:11 AM
DeMath
Quote:

Originally Posted by qzack96
I really need some help with these trig substitution problems. Please write the u and du values for each. I'd really really appreciate it !

2) Integrate [X^5 (lnx)] dx

Thank you Again !!

$\displaystyle \int {{x^5}\ln xdx} = \frac{1}{6}\int {\ln xd\left( {{x^6}} \right)} = \frac{{{x^6}}}{6}\ln x - \frac{1}{6}\int {{x^6}d\left( {\ln x} \right)} =$

$\displaystyle = \frac{{{x^6}}}{6}\ln x - \frac{1}{6}\int {\frac{{{x^6}}}{x}dx} = \frac{{{x^6}}}{6}\ln \left| x \right| - \frac{1}{6}\int {{x^5}dx} = \frac{{{x^6}}}{6}\ln \left| x \right| - \frac{{{x^6}}}{{36}} + C.$