How Do You Solve These Challenging Calculus II Integration Problems?

[The1BigJones]

This is my first posting on the site. I read the directions, I hope I follow all applicable rules. In advance, thank you for your help.

Homework Statement

Find indefinite integral:
Variables (X)
Problem:
$$\int$$Csc²(e^Cot(X))DX

Homework Equations

None:

The Attempt at a Solution

U=e^Cot(X)
DU=e^Cot(X) * -Csc²(X)
Pyth Iden Csc²(X)=1 + Tan²(X)
I keep going back to:
$$\int$$[Csc²(Something)]=-Cot(Something) I just can't figure out the relation.

Homework Statement

Find indef. Int.
Variables: X
$$\int$$ $$\frac{5}{3e^x -2}$$

Homework Equations

None

The Attempt at a Solution

U=3e^x-2
DU=3e^x

5$$\int$$$$\frac{1+3e^x -3e^x}{3e^x -2}$$
5$$\int$$$$\frac{1+3e^x}{3e^x -2}$$ - 5$$\int$$$$\frac{DU}{U}$$

Then I can't Integrate the first fraction.

Then tried conjugate:
5$$\int$$$$\frac{9e^{2x} -3e^x -2}{9e^{2x} -4}$$

Here the den. is similar to the ArcTrig integrals, however the signs are wrong and there is no sq.rt.

 

[gabbagabbahey]

Hi The1BigJones, welcome to PF!

U=e^Cot(X)
DU=e^Cot(X) * -Csc²(X)
Pyth Iden Csc²(X)=1 + Tan²(X)
I keep going back to:
$$\int$$[Csc²(Something)]=-Cot(Something) I just can't figure out the relation.

There's no need to use any trig identities here. If $u=e^{\cot x}$, then $du=-\csc^2(x) e^{\cot x} dx$ (You left out the $dx$ from your expression)...$du$ looks an awful lot like your integrand to me!

Homework Statement

Find indef. Int.
Variables: X
$$\int$$ $$\frac{5}{3e^x -2}$$

Okay, let's call this integral $I$

$$I\equiv\int\frac{5}{3e^x -2}dx$$

U=3e^x-2
DU=3e^x

5$$\int$$$$\frac{1+3e^x -3e^x}{3e^x -2}$$
5$$\int$$$$\frac{1+3e^x}{3e^x -2}$$ - 5$$\int$$$$\frac{DU}{U}$$

Then I can't Integrate the first fraction.

Hint:

$$\frac{1+3e^x}{3e^x-2}=\frac{3e^x-2+3}{3e^x-2}=1+\frac{3}{3e^x-2}$$

So now you have,

$$\int\frac{5}{3e^x -2}dx=5\int\left(1+\frac{3}{3e^x-2}\right)dx -5\ln|3e^x-2|$$

Break the integral into two pieces, and substitute $I$ in for $\int\frac{5}{3e^x -2}dx$ on both sides of the equation, and then solve for $I$ algebraically.

 

[zcd]

If you substitute $$x=\ln{t}, dx=\frac{dt}{t}$$, then you'll end up with $$\int \frac{5}{t(3t-2)}\,dt$$. From there you can apply partial fraction decomposition.