# How do you evaluate the integral int arcsin(5x-2)?

May 7, 2018

intarcsin(5x-2)dx=1/5((5x-2)arcsin(5x-2)+sqrt(1-(5x-2)²))+C, $C \in \mathbb{R}$

#### Explanation:

$\int \arcsin \left(5 x - 2\right) \mathrm{dx}$
Let $t = 5 x - 2$
$\mathrm{dt} = 5 \mathrm{dx}$
$\int \arcsin \left(5 x - 2\right) \mathrm{dx} = \frac{1}{5} \int 5 \arcsin \left(5 x - 2\right) \mathrm{dx}$
$= \frac{1}{5} \int \arcsin \left(t\right) \mathrm{dt}$
$= \frac{1}{5} \int 1 \cdot \arcsin \left(t\right) \mathrm{dt}$
Using Integration by parts :
$\int g ' \left(t\right) f \left(t\right) \mathrm{dt} = \left[g \left(t\right) f \left(t\right)\right] - \int g \left(t\right) f ' \left(t\right) \mathrm{dt}$
There:
$g ' \left(t\right) = 1$ $f \left(t\right) = \arcsin \left(t\right)$
$g \left(t\right) = t$, f'(t)=1/sqrt(1-t²)
So: 1/5intarcsin(t)dt=1/5([tarcsin(t)]-intt/sqrt(1-t²)dt)
Now let u=sqrt(1-t²)
du=-t/sqrt(1-t²)dt
So:1/5([tarcsin(t)]-intt/sqrt(1-t²)dt)=1/5([tarcsin(t)]+int1du)
$= \frac{1}{5} \left(\left[t \arcsin \left(t\right)\right] + u\right)$
=1/5(tarcsin(t)+sqrt(1-t²))
=1/5((5x-2)arcsin(5x-2)+sqrt(1-(5x-2)²))+C, $C \in \mathbb{R}$