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.