# How do you use the Fundamental Theorem of Calculus to find the derivative of int 5(sin(t))^5 dt from 5 to e^x?

Jun 25, 2016

#### Answer:

${\int}_{5}^{{e}^{x}} 5 {\sin}^{5} t \mathrm{dt} = - 5 \left\{\cos {e}^{x} - \frac{2}{3} {\cos}^{3} {e}^{x} + \frac{1}{5} {\cos}^{5} {e}^{x}\right\} + 5 \left\{\cos 5 - \frac{2}{3} {\cos}^{3} \left(5\right) + \frac{1}{5} {\cos}^{5} \left(5\right)\right\} .$

#### Explanation:

Fundamental Theorem of Calculus : $\int f \left(x\right) \mathrm{dx} = F \left(x\right) + C \Rightarrow {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\left[F \left(x\right)\right]}_{a}^{b} = F \left(b\right) - F \left(a\right)$

So, let us first find $I = \int 5 {\sin}^{5} t \mathrm{dt} = \int 5 {\sin}^{4} t \cdot \sin t \mathrm{dt} = \int 5 {\left(1 - {\cos}^{2} t\right)}^{2} \cdot \sin t \mathrm{dt} \ldots \ldots \ldots \ldots \left(1\right) .$

Now take substitution $\cos t = y ,$ so that #-sintdt=dy.

Hence, $I = 5 \int {\left(1 - {y}^{2}\right)}^{2} \cdot \left(- \mathrm{dy}\right) = - 5 \int \left(1 - 2 {y}^{2} + {y}^{4}\right) \mathrm{dy} = - 5 \left(y - \frac{2}{3} {y}^{3} + \frac{1}{5} {y}^{5}\right) + C = - 5 \left(\cos t - \frac{2}{3} {\cos}^{3} t + \frac{1}{5} {\cos}^{5} t\right) + C .$

Finally, ${\int}_{5}^{{e}^{x}} 5 {\sin}^{5} t \mathrm{dt} = - 5 \left\{\cos {e}^{x} - \frac{2}{3} {\cos}^{3} {e}^{x} + \frac{1}{5} {\cos}^{5} {e}^{x}\right\} + 5 \left\{\cos 5 - \frac{2}{3} {\cos}^{3} \left(5\right) + \frac{1}{5} {\cos}^{5} \left(5\right)\right\} .$

Jul 9, 2016

#### Answer:

$= 5 {\left(\sin \left({e}^{x}\right)\right)}^{5} \cdot {e}^{x}$

#### Explanation:

we want $\frac{d}{\mathrm{dx}} {\int}_{5}^{{e}^{x}} 5 {\left(\sin \left(t\right)\right)}^{5} \mathrm{dt}$ using FTC

FTC tells us that $\frac{d}{\mathrm{du}} {\int}_{a}^{u} f \left(t\right) \mathrm{dt} = f \left(u\right)$

here a = 5, u = e^x

so FTC + chain rules tells us that

$\frac{d}{\mathrm{dx}} {\int}_{a}^{u} f \left(t\right) \mathrm{dt} = \frac{d}{\mathrm{du}} {\int}_{a}^{u} f \left(t\right) \mathrm{dt} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= 5 {\left(\sin \left(u\right)\right)}^{5} \cdot {e}^{x}$

$= 5 {\left(\sin \left({e}^{x}\right)\right)}^{5} \cdot {e}^{x}$