# Thread: Need help understanding Fundamental theorm

1. ## Need help understanding Fundamental theorm

Why is the Integral
$\displaystyle \frac{d}{dx}\int_1^{sinx} tant dt$
equal to

$\displaystyle tan(sinx) \cdot cosx$

Why don't you have to take the the antiderivative of tan(t). Like you would if it was
$\displaystyle \frac{d}{dx}\int_1^{sinx} t+5$

Also why is the lower limit not used when solving the problem?

2. Hello swatpup32
Originally Posted by swatpup32
Why is the Integral
$\displaystyle \frac{d}{dx}\int_1^{sinx} tant dt$
equal to

$\displaystyle tan(sinx) \cdot cosx$

Why don't you have to take the the antiderivative of tan(t). Like you would if it was
$\displaystyle \frac{d}{dx}\int_1^{sinx} t+5$

Also why is the lower limit not used when solving the problem?
It doesn't matter what the function of $\displaystyle t$ is, the explanation goes like this:

$\displaystyle \frac{d}{dx}\int_1^{\sin x} f'(t)\,dt = \frac{d}{dx}\Big[f(t)\Big]_1^{\sin x}$

$\displaystyle = \frac{d}{dx}\Big(f(\sin x) - f(1)\Big)$

$\displaystyle = f'(\sin x) \times \frac{d}{dx}(\sin x) - 0$

$\displaystyle =f'(\sin x) \cos x$

OK?

Grandad

3. Originally Posted by swatpup32
Why is the Integral
$\displaystyle \frac{d}{dx}\int_1^{sinx} tant dt$
equal to

$\displaystyle tan(sinx) \cdot cosx$

Why don't you have to take the the antiderivative of tan(t). Like you would if it was
$\displaystyle \frac{d}{dx}\int_1^{sinx} t+5$

Also why is the lower limit not used when solving the problem?
You use the Fundamental Theorem of Calculus:
$\displaystyle \int_{a}^{x}f(t)dt=F(x)$
and F'(x)=f(x)

so you have
$\displaystyle (\int_{a}^{g(x)}f(t)dt)'=F(g(x))g'(x)$

For the question with the lower limit, you have the formula, from Leibinz-Newton formula:
$\displaystyle \int_{a}^{b}f(x)dx=F(b)-F(a)=-(F(a)-F(b))=-\int_{b}^{a}f(x)dx$

Hope you are familiar with Lagrange's notation.
$\displaystyle f'(x)=\frac{df}{dx}$