1. ## 2 Intergration questions

Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

1) $\int\sqrt{1 - cos(2x})sin(2x) dx$

2) $\int\tan^{-1}(3z) dz$

Thanks

2. Originally Posted by StupidIdiot
Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

1) $\int\sqrt{1 - cos(2x})sin(2x) dx$

2) $\int\tan^{-1}(3z) dz$

Thanks
for the first use a u sub

$u=1-\cos(2x) \to du=2\sin(2x)dx$

for the 2nd integrate by parts with $u=\tan^{-1}(3z)$
and $dv=dx$

This should get you started.

Good luck.

3. Hello,

Originally Posted by StupidIdiot
Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

1) $\int\sqrt{1 - cos(2x})sin(2x) dx$
Substitute : $u=1-\cos 2x \implies du=2\sin 2x dx \implies dx=\frac{du}{2 \sin 2x}$

--> $\int \sqrt{1-\cos 2x} \cdot \sin 2x dx=\int \sqrt{u} \cdot \sin 2x \cdot \frac{1}{2\sin 2x} du=\int \frac 12 \cdot \sqrt{u} du$

4. Originally Posted by StupidIdiot
Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

1) $\int\sqrt{1 - cos(2x})sin(2x) dx$

2) $\int\tan^{-1}(3z) dz$

Thanks
for the second one use integration by parts

$u=arctan(3z)$

$dv=dz$

so $\frac{3dz}{1+9z^2}$

amd $v=z$

So $\int{arctan(3z)dz}=arctan(3z)z-\int\frac{3z}{1+9z^2}dx$

$\int\frac{3z}{1+9z^2}dz=\frac{1}{6}\int\frac{18z}{ 1+9z^2}dz=\frac{\ln|1+9z^2|}{6}+C$

so $\int{arctan(3z)dz}=z\cdot{arctan(3z)}-\frac{\ln|1+9z^2|}{6}+C$

5. ## Thanks

Letting $dv=dz$ is a genius/simple step that I hadn't thought of.

Thanks to all three of you! Maybe now I can achieve my dream result (40%) tomorrow