0
$\begingroup$

Integral

I'm mostly curious as to if the way I've went about solving this is correct, or if there is a more simple way to get the answer.

So I first evaluated the top section

Step 1

And when I did that I got sin(e^x)

With that, I replaced the upper bound of the integral with sin(e^x) and integrated sin(t^2) dt

My Resulting equation was

Step 2

Which I then tried to differentiate using the product rule of three parts first being (2sin(e^x)), second sec(sin^2(e^x)), and the third tan(sin^2(e^x)). The resulting derivative I got was

enter image description here

So thats what i've done, any tips or suggestions would be appreciated mostly was hoping to get some feedback about the method I used and if its the best/easiest method for these types of problems.

$\endgroup$
  • $\begingroup$ You actually can't integrate $\sec(t^2)$ (the anti-derivative doesn't have a nice form), so the FTOC is definitely required $\endgroup$ – Dylan Dec 2 '14 at 2:34
2
$\begingroup$

Use the Fundamental Theorem of Calculus.

$$\dfrac{d}{dx} \int_a^{f(x)} G'(t) dt = \dfrac{d}{dx} \left(G(f(x))-G(a)\right) = G'(f(x))\cdot f'(x)$$

Here, $G'(t) = \sec(t^2)$ and $f(x)=\sin(e^x)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.