Question

Help with derivatives and integrals please...

Answer 1

\[g(x) = \int\limits_{1}^{x^2}t(sint)dt\]

Answer 2

we have to integrate by parts

Answer 3

hint:
\[\Large \int {t\sin t = - t\cos t - \int {\left( { - \cos t} \right)} } \;dt = ...?\]
I have integrate the factor "sint"

Answer 4

not sure where to go from there

Answer 5

here is the next step:
\[\begin{gathered}
\int {t\sin t = - t\cos t - \int {\left( { - \cos t} \right)} } \;dt = \hfill \\
\hfill \\
= - t\cos t + \int {\cos t\;dt} = ...? \hfill \\
\end{gathered} \]
please compute the remaining integral

Answer 6

what is:
\[\Large \int {\cos t\;dt} \]

Answer 7

-sint??

Answer 8

no, sint. so we have:
\[\Large g\left( x \right) = \left. { - t\cos t + \sin t} \right|_1^{{x^2}}=...?\]

Answer 9

can you tell my why we had to integrate by parts...or why it required the chain rule?

Answer 10

I have integrate by parts, since I have understood that method was right in order to solve your integral

Answer 11

came up with (-(x^2)cos(x^2) + sin(x^2)) - (-1(cos(1)) + sin(1)) , do you think I need to evaluate further?

Answer 12

no, you have to simplify only

Answer 13

can't be simplified much further, right?

Answer 14

here is a further simplification:
\[\Large \begin{gathered}
g\left( x \right) = \left. { - t\cos t + \sin t} \right|_1^{{x^2}} = \hfill \\
= - {x^2}\cos \left( {{x^2}} \right) + \sin \left( {{x^2}} \right) + \cos 1 - \sin 1 \hfill \\
\end{gathered} \]

Answer 15

thanks very much!

Answer 16

thanks!