Question

Solve.



However, I am more interested in knowing why my method does not work:


Set u=arcsinx
du=1/sqroot(1-x^2)dx
integral of (sin(u)+u) du.

Answer 1

Your method is fine, but your result is incorrect. You made a mistake with computing \(\int u~du\)

Answer 2

So where you have \(\int (\sin u+u)~du\), you should have \(-\cos u+\dfrac{u^2}{2}+C\)

Answer 3

I corrected it. :) But, when I graph the result from my teacher and my result, the slopes are different.

Answer 4

Yes, that is what I got.

Answer 5

Right, so back-substituting yields \(-\cos(\arcsin x)+\dfrac{\arcsin^2x}{2}+C\), which is the same as your teacher's result.

Answer 6

You are a genius.

Answer 7

I get it all now. THANKS. :)

Answer 8

yw!