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 !!

2. Originally Posted by qzack96

3) Integrate [sinx*cosx] dx
$\int \sin(x)\cos(x) dx$

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

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

$\int u~du$

can you take it from here?

3. 1 let $u=e^x$, so your integral is now

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

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

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

The last way makes this

${1\over 2}\int \sin (2x)dx={-1\over 4} \cos (2x)+C$

4. 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?

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

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