# Evaluate the following integral $\int_{0}^{10}\sqrt{-175e^{-t/4}+400}dt$

I am trying to evalute the following integral:
$$\int_{0}^{10}\sqrt{-175e^{-t/4}+400}dt$$
Any suggestions how to start?
Thanks!

#### Solutions Collecting From Web of "Evaluate the following integral $\int_{0}^{10}\sqrt{-175e^{-t/4}+400}dt$"

Make the substitution

$$u=400-175 e^{-t/4} \implies t = -4 \log{\frac{400-u}{175}} \implies dt = \frac{4}{400-u} du$$

The integral is then equal to

$$4 \int_{225}^{400-175 e^{-5/2}} \frac{du}{400-u} \sqrt{u}$$

Then make the substitution $u=v^2$ to get a more familiar form:

$$8 \int_{15}^{\sqrt{400-175 e^{-5/2}}} dv \frac{v^2}{400-v^2} = 8 \int_{15}^{\sqrt{400-175 e^{-5/2}}} dv \left (\frac{400}{400-v^2} – 1 \right )$$

Use the fact that

$$\int \frac{dy}{a^2-y^2} = \frac12 \log{\frac{a+y}{a-y}}$$

and the rest is arithmetic.