# How do you use Part 1 of the fundamental theorem of calculus to find the derivative of the function y= int (1+v^2)^10 dv from sinx to cosx? Please help have been looking at problem for a hour can't figure out how to work it?

May 28, 2018

$\frac{d}{\mathrm{dx}} \left[{\int}_{\sin x}^{\cos x} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv}\right] = \sin x + \cos x - {\left(1 + {\cos}^{2} x\right)}^{10} \sin x - {\left(1 + {\sin}^{2} x\right)}^{10} \cos x$

#### Explanation:

Let:

$F \left(u\right) = {\int}_{0}^{u} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv}$

based on the fundamental theorem of calculus:

$\frac{\mathrm{dF}}{\mathrm{du}} = {\left(1 + {u}^{2}\right)}^{10} - 1$

Now:

${\int}_{\sin x}^{\cos x} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv} = F \left(\cos x\right) - F \left(\sin x\right)$

and using the linearity of the derivative and the chain rule:

$\frac{d}{\mathrm{dx}} \left[{\int}_{\sin x}^{\cos x} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv}\right] = F ' \left(\cos x\right) \frac{d}{\mathrm{dx}} \cos x - F ' \left(\sin x\right) \frac{d}{\mathrm{dx}} \sin x$

$\frac{d}{\mathrm{dx}} \left[{\int}_{\sin x}^{\cos x} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv}\right] = - \left({\left(1 + {\cos}^{2} x\right)}^{10} - 1\right) \sin x - \left({\left(1 + {\sin}^{2} x\right)}^{10} - 1\right) \cos x$

$\frac{d}{\mathrm{dx}} \left[{\int}_{\sin x}^{\cos x} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv}\right] = \sin x + \cos x - {\left(1 + {\cos}^{2} x\right)}^{10} \sin x - {\left(1 + {\sin}^{2} x\right)}^{10} \cos x$

May 28, 2018

See below

#### Explanation:

FTC, Part 1 states, broadly, that if:

• $F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}$

then:

• $F ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$

This can then be modified for an interval that includes a function $u = u \left(x\right)$, by using the chain rule:

• $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} \implies \frac{d}{\mathrm{dx}} {\int}_{a}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt} = f \left(u \left(x\right)\right) \cdot u ' \left(x\right)$

Then by manipulating the interval, you can get a complete result:

• ${\int}_{v \left(x\right)}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt} = {\int}_{0}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt} - {\int}_{0}^{v \left(x\right)} f \left(t\right) \setminus \mathrm{dt}$

• $\implies \frac{d}{\mathrm{dx}} {\int}_{v \left(x\right)}^{u \left(x\right)} f \left(t\right) \setminus \mathrm{dt} = f \left(u \left(x\right)\right) \cdot u ' \left(x\right) - f \left(v \left(x\right)\right) \cdot v ' \left(x\right)$

In this case, $v$ is the dummy variable instead of $t$. Matters not.

$\frac{d}{\mathrm{dx}} {\int}_{\sin x}^{\cos x} {\left(1 + {v}^{2}\right)}^{10} \mathrm{dv}$

$= {\left(1 + {\cos}^{2} x\right)}^{10} \cdot \left(- \sin x\right) - {\left(1 + {\sin}^{2} x\right)}^{10} \cdot \cos x$

....which might simplify a little