1. More Integrals

1. $xsin(x^2)cos(3x^2)dx$

2.

2. For the first one, let $u=x^2.$ After turning the resultant product into a sum the rest follows.

For the second one, let $u=x\ln x.$ The remaining integral can be killed with a trig. or hyperbolic substitution.

3. Originally Posted by Krizalid
For the first one, let $u=x^2.$ After turning the resultant product into a sum the rest follows.

For the second one, let $u=x\ln x.$ The remaining integral can be killed with a trig. or hyperbolic substitution.
I got the first one, but I'm still having some trouble with the second one. I'm not really understanding it.

4. Hello, FalconPUNCH!!

Krizalid is absolutely correct . . .

$\int(1+\ln x)\sqrt{1+(x\ln x)^2}\,dx$

Let $u \:=\:x\ln x\quad\Rightarrow\quad du \:=\:\left(x\!\cdot\!\frac{1}{x} + \ln x\right)\,dx\:=\1 + \ln x)\,dx" alt="u \:=\:x\ln x\quad\Rightarrow\quad du \:=\:\left(x\!\cdot\!\frac{1}{x} + \ln x\right)\,dx\:=\1 + \ln x)\,dx" />

We have: . $\int\underbrace{\sqrt{1 + (x\ln x)^2}}\underbrace{(1 + \ln x)dx}$
. . . . . . . . . . . . . $\downarrow\qquad\qquad\swarrow$
. . . . . . . . $= \;\int \sqrt{1 + u^2}\;du$

Now let $u \:= \:\tan\theta\quad\Rightarrow\quad du \:=\:\sec^2\!\theta\,d\theta \quad\hdots\quad \text{etc.}$