Question

d/dx*integral of sqrt(t^2+1)dt from sinx to cosx

Answer 1

\[ \Large\frac{d}{dx} \int \sqrt{t^2+1} dt\]
correct?

Answer 2

yes... sinx to cosx are the bounds

Answer 3

Basically you compute the the integral using the fundamental theorem and then derive it:
\[\Large \int_{\cos x}^{\sin x} \sqrt{t^2+1}dt=F(\sin(x))-F(\cos(x)) \]

Answer 4

now derive both sides, hint: use the chain rule

Answer 5

Just to write out the differentiation:
\[\Large \left(\frac{d}{dx} \int_{\cos x}^{\sin x} \sqrt{t^2+1}dt\right)=\frac{d}{dx}\left(F(\sin(x))-F(\cos(x))\right)\]

Answer 6

so F = sqrt(t^2+1)? or am i wrong?

Answer 7

differentiate sqrt(sin^2x+1)-sqrt(cos^2x+1)...?

Answer 8

no you different the RHS, do you know how to do that?

You might want to remember the relationship that:
\[\Large F'(x)=f(x) \]

Answer 9

so F(cosx)+F(sinx) is the differentiation?

Answer 10

hehe, no that's still not quite what it is, as I mentioned above, you need the chain rule, let me try and help you with the first term, I exclude the second one so you can do it for yourself:
\[\Large \frac{d}{dx}F(\sin(x))=\left(F(\sin(x)\right)'=f(\sin(x))\cdot \cos(x) \]
See and verify for yourself that this is nothing but the chain rule, you can also do that by substitution if you like. Substitute u=sin(x) for instance

Answer 11

so d/dx F(sin(x))-F(cos(x)) = f(sin(x))*cosx + f(cos(x))*sin(x)??

Answer 12

there you go

Answer 13

so how do i write the answer correct i suppose?

Answer 14

well you substitute everything back,
\[\Large f(t)=\sqrt{t^2+1} \]
so what is \[\Large f(\sin(x))=? \]

Answer 15

\[d/dx \int\limits_{cosx}^{sinx} \sqrt{t^{2}+1}dt = \sqrt{\sin^{2}x +1}*cosx + \sqrt{\cos^{2}x+1}*sinx\]

Answer 16

correct??

Answer 17

that's it!

Answer 18

appreciate it!

Answer 19

you're welcome