1. ## I don't like Integrals. Help PLEASE

Hi again, guys. Tomorrow I will be getting some tutoring help with Integrals so I'm hoping you guys help me out with 2 integral questions. Please and thank you! <3

Here are the problems:

$\int 8csc(8x-9)dx$

and

$\int^{\frac{2\pi}{3}}_0 tan(\frac{x}{2})dx$

Hi again, guys. Tomorrow I will be getting some tutoring help with Integrals so I'm hoping you guys help me out with 2 integral questions. Please and thank you! <3

Here are the problems:

$\int 8csc(8x-9)dx$

and

$\int^{\frac{2\pi}{3}}_0 tan(\frac{x}{2})dx$

For the first one, try u = 8x-9...You end up with csc(u). Simply by multiplying $\frac{\csc{u}+\cot{u}}{\csc{u}+\cot{u}}$ with $\csc{u}$ you get an integral of the form:
$\int \frac{f'(x)}{f(x)}~dx$

You see that this is a ln antiderivative form right away.

As for the second one, simply multiply by $\frac{\sec{\frac{x}{2}}}{\sec{\frac{x}{2}}}$...Against, this is a ln antiderivative.

3. Originally Posted by Chop Suey
As for the second one, simply multiply by $\frac{\sec{\frac{x}{2}}}{\sec{\frac{x}{2}}}$...Against, this is a ln antiderivative.
You can make it a tad easier by applying the definition of tangent...because I would not have thought to multiply through by $\frac{\sec\left(\frac{x}{2}\right)}{\sec\left(\fra c{x}{2}\right)}$ [you'll end up with the same thing anyway]

$\tan\left(\frac{x}{2}\right)=\frac{\sin\left(\disp laystyle\frac{x}{2}\right)}{\cos\left(\displaystyl e\frac{x}{2}\right)}$

Then the u-substitution becomes obvious....

--Chris

4. Originally Posted by Chris L T521
You can make it a tad easier by applying the definition of tangent
I know, but I just thought I'd do things differently every once in a while.