Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of
h(x) = \int_{-1}^{\sin(x)} (\cos(t^2)+t)\; dt

Answer 1

\[h(x) = \int_{-1}^{\sin(x)} (\cos(t^2)+t)\; dt\]

Answer 2

-2cosx(cos(sin^2x))+sinx my answer. Thanks ash for puting up the right format. I can't figureout how my answer is wrong though

Answer 3

You put t= sin x and multiply the expression by the derivative of sin x

Answer 4

that's what I did but somehow I got the wrong answer. could you look at my answer and tell me why it's wrong?

Answer 5

t= sin x, as the lower limit is constant it'd be 0
\[\{\frac{d}{dx} (\sin x) \}\times ( \cos(\sin^2 x))+\sin x )-0\]
\[\cos x ( \cos(\sin^2 x))+\sin x )\]

Answer 6

\[ \cos x(\cos(\sin^2 x)+\sin x )\]

Answer 7

Did you get it?

Answer 8

I see where I went wrong thanks!