How Do You Solve These Challenging Calculus II Integration Problems?

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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:
\intCsc2(eCot(X))DX

Homework Equations


None:

The Attempt at a Solution


U=eCot(X)
DU=eCot(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=3ex-2
DU=3ex

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.
 
Last edited:
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Hi The1BigJones, welcome to PF!

The1BigJones said:
U=eCot(X)
DU=eCot(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!:wink:

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=3ex-2
DU=3ex

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.
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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