Use part 1 of the Fundamental Theorem of Calculus to find the derivative.

1. Jan 13, 2013

Lo.Lee.Ta.

1. h(x) = ∫-3 to sin(x) of (cos(t^3) + t)dt

2. Okay, I know that you are supposed to replace t with the upper limit, and then I think you multiply that term by the derivative of the upper limit.

So I thought it would be:

cos(sinx)^3 * cos(x) + sinxcosx

But what even is cos(sinx)? I've seen sinxcosx before but not that...
But I put this answer anyway into my computer hw system, and it's counted wrong.

Would you please tell me what I'm doing wrong?
Thank you! :)

2. Jan 13, 2013

Dick

(cos(sin(x)^3)+sin(x))*cos(x) is not the same thing as ((cos(sin(x))^3+sin(x))*cos(x). Do you see the difference? Which would you say is correct? I suspect you aren't using enough parentheses.

Last edited: Jan 13, 2013
3. Jan 13, 2013

Dick

I'll try to use some amateurish tex here to try to emphasize the difference between the important parts. $\cos((\sin x)^3)$ vs $(\cos(\sin x))^3$ Which one do you want?

4. Jan 13, 2013

Lo.Lee.Ta.

You were absolutely right! The parentheses were off!

cos((sinx)^3) should be the right one. I only want to cube the sinx.

[cos(sin(x)^3) + sin(x)]cos(x)

Oh, this may sound dumb, but... if you were to multiply this out, how would it be written?

Could you condense the cosines (to something like cos^2(x)(sin(x)^3) or would it have to remain as cos(sin(x)^3)*cos(x) ?

cos^2(x) means cos(x)*cos(x)... But there are not two x's... So is cos(sin(x)^3)*cos(x) the most simplified version?

Thank you so much for helping! :)

5. Jan 13, 2013

Dick

cos(sin(x)^3) is the most simplified version. There's not much you can do with it. There's no trig law that will let you expand it out.

Last edited: Jan 13, 2013
6. Jan 13, 2013

Lo.Lee.Ta.

Oh, okay. Thank you so much! :)