# Thread: a few more integrals from this year's MIT Integration Bee

1. ## a few more integrals from this year's MIT Integration Bee

Since I don't know the solutions to all of these, I'm posting them here instead of in the Math Challenge Problems forum.

None of these integrals should have very complicated solutions.

EDIT: The fourth one might be too easy.

1) $\int^{\pi/2}_{0} \sin (x) \sin (2x) \sin (3x) \ dx$

2) $\int x \ln \Big( 1 + \frac{1}{x} \Big) \ dx$

3) $\int^{1}_{0} \sin^{2} (\ln x) \ dx$

4) $\int x^{x} ( 1 + \ln x ) \ dx$

5) $\int^{2}_{1} (x-1)^{1/2} (2-x)^{1/2} \ dx$

2. $I=\int x \ln(1+\frac{1}{x})dx=\int \left(\frac{x^2}{2}\right)' \ln\left(1+\frac{1}{x}\right)dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \ln\left(1+\frac{1}{x}\right) \right)'dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \dfrac{-1}{x(x+1)} \right)dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\int \left(\frac{x}{2(x+1)}\right)dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra c{1}{2}\int \dfrac{x+1-1}{x+1}dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra c{1}{2}\left(\int 1dx -\int\dfrac{1}{x+1}dx \right)$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra c{1}{2}\left(x-\ln(1+x)\right) + C$

I think that it's ok

3. The fourth:

$I=\int x^x (1+\ln(x))dx = \int e^{x\ln(x)}(1+\ln(x))dx$

Let be $x\ln(x)=u$, then

$I=e^u du = x^x+C$

4. Originally Posted by felper
$I=\int x \ln(1+\frac{1}{x})dx=\int \left(\frac{x^2}{2}\right)' \ln\left(1+\frac{1}{x}\right)dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \ln\left(1+\frac{1}{x}\right) \right)'dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \dfrac{-1}{x(x+1)} \right)dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\int \left(\frac{x}{2(x+1)}\right)dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra c{1}{2}\int \dfrac{x+1-1}{x+1}dx$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra c{1}{2}\left(\int 1dx -\int\dfrac{1}{x+1}dx \right)$
$I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra c{1}{2}\left(x-\ln(1+x)\right) + C$

I think that it's ok
Wake up! You're correct.

5. 4) $\int x^{x} ( 1 + \ln x ) \ dx$

=

$\int e^{xlnx} ( 1 + \ln x ) \ dx$

let u = $xlnx$

$\int e^u du$ = $x^x+C$

6. Originally Posted by felper
The fourth:

$I=\int x^x (1+\ln(x))dx = \int e^{x\ln(x)}(1+\ln(x))dx$

Let be $x\ln(x)=u$, then

$I=e^u du = x^x+C$
LOL...this one was way too easy...not inclined to do any with bounds...no indefinites left. i would have posted this sooner, probably a minute after you, but i took a trip to the can.

7. $I=\int_0^1 \sin^2(\ln(x))dx=\int_0^1 \dfrac{1-\cos(2\ln(x))}{2}dx$
$I=\dfrac{1}{2}-\int_0^1 \dfrac{\cos(2\ln(x))}{2}dx$
Let be $\ln(x)=u \Leftrightarrow e^u=x \implies dx=e^u du$
$\int_0^1 \dfrac{\cos(2\ln(x))}{2}dx=\dfrac{1}{2}\int_{-\infty}^0\cos(2u)e^u du$
Let be $u=-t \implies du=-dt$ then:
$\int_0^1 \dfrac{\cos(2\ln(x))}{2}dx=-\dfrac{1}{2}\int_{\infty}^0\cos(-2u)e^{-u} du=\dfrac{1}{2}\int_{0}^{\infty}\cos(-2u)e^{-u} du$

Which is the Laplace Transform for cos(2t), so, the integral is

$I=\frac{1}{2}-\frac{1}{2}\cdot \frac{1}{5}=\frac{2}{5}$

End.

8. $I=\int^{\pi/2}_{0} \sin (x) \sin (2x) \sin (3x) \ dx=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}(\cos(2x)-\cos(4x))\sin(2x)dx
$

$I=\frac{1}{2}\left (\int_0^{\frac{\pi}{2}}\cos(2x)\sin(2x)dx-\int_0^{\frac{\pi}{2}}\cos(4x)\sin(2x)dx \right)$
$I=\frac{1}{2}\left( \frac{1}{2}\int_0^{\frac{\pi}{2}}\sin(4x)dx - \frac{1}{2}\int_0^{\frac{\pi}{2}}(\sin(6x)-\sin(2x))dx \right)$
$I=\frac{1}{4}\left(\int_0^{\frac{\pi}{2}}\sin(4x)d x+\int_0^{\frac{\pi}{2}}\sin(2x)dx-\int_0^{\frac{\pi}{2}}\sin(6x)dx \right)$
$I=\frac{1}{4}\left( 0+1-\frac{1}{3}\right)=\frac{1}{6}$

9. Originally Posted by felper
$I=\int^{\pi/2}_{0} \sin (x) \sin (2x) \sin (3x) \ dx=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}(\cos(2x)-\cos(4x))\sin(2x)dx
$

$I=\frac{1}{2}\left (\int_0^{\frac{\pi}{2}}\cos(2x)\sin(2x)dx-\int_0^{\frac{\pi}{2}}\cos(4x)\sin(2x)dx \right)$
$I=\frac{1}{2}\left( \frac{1}{2}\int_0^{\frac{\pi}{2}}\sin(4x)dx - \frac{1}{2}\int_0^{\frac{\pi}{2}}(\sin(6x)-\sin(2x))dx \right)$
$I=\frac{1}{4}\left(\int_0^{\frac{\pi}{2}}\sin(4x)d x+\int_0^{\frac{\pi}{2}}\sin(2x)dx-\int_0^{\frac{\pi}{2}}\sin(6x)dx \right)$
$I=\frac{1}{4}\left( 0+1-\frac{1}{3}\right)=\frac{1}{6}$
Correct again. These are too easy for you.

10. Do you have a website for thier problems?

11. $
\int^{2}_{1} (x-1)^{1/2} (2-x)^{1/2} \ dx
$

Choose

$u=x-1$

$\Rightarrow$

$\int^{2}_{1} (x-1)^{1/2} (2-x)^{1/2} \ dx$

=

$\int^{1}_{0} (u)^{1/2} (1-u)^{1/2} \ du$

Choose

$sin^2v = u$

$du = 2sinvcosv dv = sin2v dv$

$\Rightarrow$

$\int^{1}_{0} (u)^{1/2} (1-u)^{1/2} \ du$

=

$\int^{\pi/2}_{0} (sinv)(cos^2v)^{1/2} \ sin2v \ dv$

=

$\int^{\pi/2}_{0} sinv \ sin2v \ cosv \ dv$

=

$\int^{\pi/2}_{0}cosv\times\frac{1}{2}(cosv-cos3v)dv$

=

$\frac{1}{2}\int^{\pi/2}_{0}cos^2v-\frac{1}{2}(cos2v+cos4v) \ dv$

The last 2 terms evaluate to 0.

$\int^{\pi/2}_{0} \frac{1}{2} \times\cos^2v \ dv$

=

$\pi/8$

All in all, quite a yawner of a set. The one where Felper recognized the Laplace was the best one, but even that one was str8tforward. This must be for the total newbie undergrads at mit.