Integration By Parts: int (arcsin x) dx

[Annienakki]

Hi Guys,

My name is Annie, I just joined the forum yesterday. I'm in Calc. II right now and loving it! So, I thought I'd say hello by asking a question. But the funny thing is, as I was getting ready to type in the integral, I figured out what the book did, lol.

Anyway, here's what I was going to ask before I figured out what they did:

I'm doing practice problems in Schaum's 3000 Solved Problems In Calculus, Chapter 28, Integration By Parts. #28.4 is:

?arcsin x dx

I let u=arcsin x and dv= dx giving me

du=1/(1-x^2)^1/2 dx and v=x....the formula now looks like this:

xarcsinx - ?x/(1-x^2)^1/2 dx

This is where the book lost me...their next step has the integral part as 1/2?(1-x)^-1/2 * (-2x)dx

Well, yesterday I was stumped how they got there, but as I said previously, as I was getting ready to post I was able to see what they did...they multiplied the numerator by -2 and divided the integral by -2 to clear it...duh...but wait, I know it's early here (I'm in Tucson and it's 4:20) but now I can't see how they got the next step, lol! Hmmmm well I will write their next step and then go see if I can solve it....

Here's their next step:

xarcsinx + 1/2 * 2(1-x^2)^1/2 + C .....How did they clear that integral? Brain dead this morning....

Thanks, guys....

Annie

 

[royhaas]

Re: Integration By Parts

Look at the the derivative of \(\displaystyle \sqrt{1-x^2}\). Don't forget the chain rule.

 

[Annienakki]

Re: Integration By Parts

Thanks, Roy.... I get it! I need to make a note to myself to remember things like the chain rule when I'm doing tougher integrals...I tend to focus on trig rules, substitution, etc. and forget about earlier techniques like the chain rule...

Thanks again,

Annie