Basic question about integrating by parts

Question

Suppose I want to solve the following:

$\int \arcsin(t) \space dt=?$

In order to solve this I would use integration by parts:

$\int \ uv'dx=uv-\int \ u'v \space dx$

If I let $v'=\arcsin(t)$ then obviously I need to know the antiderivative of $v'$ in order to integrate by parts. However, this assumes that I already know the antiderivative of that function. Can I even integrate if I don't know the antiderivative of one of my functions?

Thanks.

Answer 1

The whole point of integration by parts is to avoid integrating the function if you can avoid it: in your example, you don't know the antiderivative of $\arcsin$, but you do know the antiderivative of $1$, which is $x$. Therefore you take $v'=1$ and get $$ \int \arcsin{x} \, dx = x \arcsin{x} - \int x arcsin'{x} \, dx \\ = x\arcsin{x} - \int \frac{x}{\sqrt{1-x^2}} \, dx. $$ This is in the form $f'(x)[f(x)]^{-1/2}$, so can be integrated easily to get $$ \int \arcsin{x} \, dx = x\arcsin{x} +\sqrt{1-x^2}. $$

The same trick works for the other inverse trigonometric functions, and $\log{x}$.

Answer 2

$$\int\arcsin(t)\ dt$$

$$=\arcsin(t)\int\ dt-\int\left[\dfrac{d[\arcsin(t)]}{dt}\int\ dt\right]dt$$

$$=t\arcsin(t)-\int\dfrac t{\sqrt{1-t^2}}\ dt$$

Set $\sqrt{1-t^2}=y$