integral of sinxcosx

Can someone confirm to me if i am doing this correctly. I did it by substitution and came up with 1/2 sin(sq)x. However since then i realised it is easier to use the identity 2sinxcosx =sin2x and that gives an answer of
-1/4 cos2x.
So basically can someone tell me if 1/2sinsqx is equal to -1/4cos2x, if it isn't come someone go trough how you would find the integral by substitution using u=sinx

cheers

Reply 1

silent ninja

both, either.

Reply 2

AlphaNumeric

Try putting in a value of x, such as x=0. If you were right, you'd have just proved 0=1 :wink:

When you use u=sinx, you get the integral

\int \sin x \cos x dx = \int u\sqrt{1-u^{2}}\sqrt{1-u^{2}}du = \int u-u^{3} du

Integrating that and subbing back in for u doesn't involve

\sin \sqrt{x}

at all.

Reply 3

Meat Loaf Rocks

AlphaNumeric

Try putting in a value of x, such as x=0. If you were right, you'd have just proved 0=1 :wink:

When you use u=sinx, you get the integral

\int \sin x \cos x dx = \int u\sqrt{1-u^{2}}\sqrt{1-u^{2}}du = \int u-u^{3} du

Integrating that and subbing back in for u doesn't involve

\sin \sqrt{x}

at all.

when i wrote sin(sq)x I meant sin sqaured x not square root, sorry i should have been clearer

cheers guys

Reply 4

silent ninja

but differentiating 1/2sin²x= sinxcosx
and -1/4 cos2x = 1/2sin2x = sinxcos

?

I had this in a question before and used both just to see if it gave the same answers, and it did.

edit: i think you mistaken the ops typing for squ as √ but he meant squared