1. ## 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

2. 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?

3. 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

$u=sinx$

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

Nevermind, looks like you edited your post.

4. 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?

5. $\int \sin{x} \cos{x} \, dx$

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

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

can you get the textbook solution from this?

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

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

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

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.