# trig integration

• Jul 5th 2009, 03:38 PM
slaypullingcat
trig integration
Hi, I can't integrate sin x cos x. I tried a u substitution with u=sinx and dx=cos x du and got 1/2(sin x)^2+c. My text book answer is -1/4cos 2x+c.

Can someone please show where I am going wrong?

Is my u substitution wrong?

Thank you
• Jul 5th 2009, 03:52 PM
skeeter
Quote:

Originally Posted by slaypullingcat
Hi, I can't integrate sin x cos x. I tried a u substitution with u=sinx and dx=cos x du and got 1/2(sin x)^2+c. My text book answer is -1/4cos 2x+c.

Can someone please show where I am going wrong?

Is my u substitution wrong?

Thank you

you are not "wrong".

try graphing your solution and the book's solution ... what do you notice?
• Jul 5th 2009, 03:53 PM
VonNemo19
Quote:

Originally Posted by slaypullingcat
Hi, I can't integrate sin x cos x. I tried a u substitution with u=sinx and dx=cos x du and got 1/2(sin x)^2+c. My text book answer is -1/4cos 2x+c.

Can someone please show where I am going wrong?

Is my u substitution wrong?

Thank you

$\displaystyle u=sinx$

$\displaystyle du=cosx\cdot{dx}\Rightarrow{dx}=\frac{du}{cosx}$ Right? I don't know, I'm asking.

Nevermind, looks like you edited your post.
• Jul 5th 2009, 04:00 PM
slaypullingcat
Quote:

Originally Posted by skeeter
you are not "wrong".

try graphing your solution and the book's solution ... what do you notice?

Thank you for that, but I still don't know how to integrate the 'text book' way. Can you please even suggest the method they have used or a different approach I should try?
• Jul 5th 2009, 04:15 PM
skeeter
$\displaystyle \int \sin{x} \cos{x} \, dx$

$\displaystyle \frac{1}{2} \int 2\sin{x} \cos{x} \, dx$

$\displaystyle \frac{1}{2} \int \sin(2x) \, dx$

can you get the textbook solution from this?
• Jul 5th 2009, 05:49 PM
slaypullingcat
Quote:

Originally Posted by skeeter
$\displaystyle \int \sin{x} \cos{x} \, dx$

$\displaystyle \frac{1}{2} \int 2\sin{x} \cos{x} \, dx$

$\displaystyle \frac{1}{2} \int \sin(2x) \, dx$

can you get the textbook solution from this?

Yes I can find it now. I am not very good at recognising trig identities. Thank you for your post, it makes sense now. Cheers.