Trigonmentric Integral

[simplybilly]

EDIT: I am pretty sure I message up when I cancelled out stuff by leaving cos(x) and sin(x) instead of making them 1 / sin(x) and 1 / cos(x)

Alright, so I have been trying to figure out this problem and the solution I get is incorrect

The problem (int) is suppose to be the integral symbol

(int) 3 [ (cos(x) + sin(x)) / sin (2x) ] dx

I use sin(2x) = 2sin(x)cos(x) which gives me

3 (int) [ (cos(x) + sin(x)) / (2sin(x)cos(x)) ] dx

I then take out 1/2 and split

3/2 (int) [ (cos(x) / sin(x)) + (sin(x) / cos(x)) ] dx

Then I use some trig identities to get

3/2 (int) [ cot(x) + tan(x) ] dx

Then some integral identities to get

3/2 [ ln|sin(x)| -ln|cos(x)| ]

Then just multiply through to get

3/2ln|sin(x)| - 3/2ln|cos(x)| + C

So what am I doing wrong?

 

[srmichael]

EDIT: I am pretty sure I message up when I cancelled out stuff by leaving cos(x) and sin(x) instead of making them 1 / sin(x) and 1 / cos(x)

Alright, so I have been trying to figure out this problem and the solution I get is incorrect

The problem (int) is suppose to be the integral symbol

(int) 3 [ (cos(x) + sin(x)) / sin (2x) ] dx

I use sin(2x) = 2sin(x)cos(x) which gives me

3 (int) [ (cos(x) + sin(x)) / (2sin(x)cos(x)) ] dx

I then take out 1/2 and split

3/2 (int) [ (cos(x) / sin(x)) + (sin(x) / cos(x)) ] dx ===> Incorrect. See below
Then I use some trig identities to get

3/2 (int) [ cot(x) + tan(x) ] dx

Then some integral identities to get

3/2 [ ln|sin(x)| -ln|cos(x)| ]

Then just multiply through to get

3/2ln|sin(x)| - 3/2ln|cos(x)| + C

So what am I doing wrong?

\(\displaystyle 3\int[\frac{\cos(x)+\sin(x)}{2\sin(x)\cos(x)}]dx=\frac{3}{2}\int[\frac{\cos(x)}{\sin(x)\cos(x)}+\frac{\sin(x)}{\sin(x)\cos(x)}]dx=\frac{3}{2}\int[\frac{1}{\sin(x)}+\frac{1}{\cos(x)}]dx=\frac{3}{2}\int[\csc(x)+\sec(x)]dx\)

Now proceed....