Integrating Fractions: Understanding How to Integrate 2+3sin^2x/5sin^2x

[Cmertin]

I'm having some problems integrating fractions. If you could help me understand it, that would be great.

Homework Statement

\int\frac{2+3sin^{2}x}{5sin^{2}x}

Homework Equations

\int(x)dx=\frac{x^{n+1}}{n+1}

The Attempt at a Solution

\frac{2+3sin^{2}x}{5sin^{2}x}=\frac{1}{5}(\frac{2}{sin^{2}x}+\frac{3sin^{2}x}{sin^{2}x})

=\frac{1}{5}(\frac{2}{sin^{2}x}+\frac{3sin^{2}x}{sin^{2}x})

=\frac{1}{5}(\frac{2}{sin^{2}x}+3)

\frac{1}{5}\int\frac{2}{sin^{2}x}+3 dx=\frac{1}{5}(\frac{-2}{sin(x)cos(x)}+3x)+C

This is wrong though because the answer is supposed to be:
\frac{1}{5}(3x-2cot(x))

What did I do wrong?

 

[vela]

How did you go from

\int \frac{1}{\sin^2 x}\,dx

to

-\frac{1}{\sin x\cos x}

 

[Cmertin]

How did you go from

\int \frac{1}{\sin^2 x}\,dx

to

-\frac{1}{\sin x\cos x}

Actually, now that I look at it that doesn't make sense. But I'm still stuck and can't figure out the steps before the answer...

\frac{1}{5}\int\frac{2}{sin^{2}x}+3 dx=?

 

[vela]

Hint: 1/sin x = csc x.

 

[Cmertin]

Hint: 1/sin x = csc x.

\frac{1}{5}\int\frac{2}{sin^{2}x}+3 dx

=\frac{1}{5}(3x+\int\frac{2}{sin^{2}x}dx)

=\frac{1}{5}(3x+2csc^{2}x)But the answer is:
\frac{1}{5}(3x-2cot(x))
Found here

I have no idea where they got cotan from.

 

[Mentallic]

\frac{1}{5}\int\frac{2}{sin^{2}x}+3 dx

=\frac{1}{5}(3x+\int\frac{2}{sin^{2}x}dx)

=\frac{1}{5}(3x+2csc^{2}x)But the answer is:
\frac{1}{5}(3x-2cot(x))
Found here

I have no idea where they got cotan from.

You never took the integral of the csc²(x). Take the derivative of tan(x) and see what you get.

 

[Mark44]

To elaborate a bit on what Mentallic said, this is what you did:
\int \frac{2}{sin^2(x)}dx~=~\int 2~csc^2(x) dx~=~2csc^2(x)

If only integration were that simple!

 

[Cmertin]

You never took the integral of the csc²(x). Take the derivative of tan(x) and see what you get.

Ah, the integral of csc²(x) is -cot(x). That's where my mistake was. Thanks.

 

[Cmertin]

To elaborate a bit on what Mentallic said, this is what you did:
\int \frac{2}{sin^2(x)}dx~=~\int 2~csc^2(x) dx~=~2csc^2(x)

If only integration were that simple!

Yea, I wish it was that simple :P Thanks for the clarification.