# Having trouble finding the integral here. ?

Having a tough time with this problem. Can’t even figure out the right way to use the hint. Lol

### 2 Answers

- 1 month agoFavourite answer
e^(x) = u

e^(x) * dx = du

5 * e^(x) * dx / sqrt(4 - 9 * e^(2x)) =>

5 * du / sqrt(4 - 9 * u^2)

Recall that 1 - sin(t)^2 = cos(t)^2

4 - 9 * u^2 = 4 - 4 * sin(t)^2

-9 * u^2 = -4 * sin(t)^2

u^2 = (4/9) * sin(t)^2

u = (2/3) * sin(t)

du = (2/3) * cos(t) * dt

5 * du / sqrt(4 - 9 * u^2) =>

5 * (2/3) * cos(t) * dt / sqrt(4 - 4 * sin(t)^2) =>

(10/3) * cos(t) * dt / sqrt(4 * cos(t)^2) =>

(10/3) * cos(t) * dt / (2 * cos(t)) =>

(5/3) * dt

Integrate

(5/3) * t + C

u = (2/3) * sin(t)

(3/2) * u = sin(t)

arcsin(3u / 2) = t

(5/3) * arcsin(3u / 2) + C

u = e^(x)

(5/3) * arcsin(3 * e^(x) / 2) + C

Truth be told, we could have used one substitution, but I find it's oftentimes easier to express an expression in terms of a variable and then express that variable in terms of another expression instead of substituting one expression for another. It's less of a conceptual leap.

4 - 9 * e^(2x) = 4 - 4 * sin(t)^2

9 * e^(2x) = 4 * sin(t)^2

3 * e^(x) = 2 * sin(t)

e^(x) = (2/3) * sin(t)

e^(x) * dx = (2/3) * cos(t) * dt

5 * e^(x) * dx / sqrt(4 - 9 * e^(2x)) =>

5 * (2/3) * cos(t) * dt / sqrt(4 - 4 * sin(t)^2) =>

(10/3) * cos(t) * dt / (2 * cos(t)) =>

(5/3) * dt

(5/3) * t + C

e^(x) = (2/3) * sin(t)

(3 * e^(x) / 2) = sin(t)

t = arcsin(3 * e^(x) / 2)

(5/3) * arcsin(3 * e^(x) / 2) + C

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- Ian HLv 71 month ago
d/dy[arcsin(y/2)] = 1/√[4 – y^2] let y = (3/2)e^x

d/dx{arcsin[(3/2)e^x]} = 3e^x/√[4 – (3e^x)^2]

I = ∫[5e^x/√[4 – 9e^(2x)]dx = (5/3)arcsin[(3/2)e^x]

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