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.

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Yes, that is what I got.

anonymous · Answer

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

anonymous · Answer

You are a genius.

anonymous · Answer

I get it all now. THANKS. :)

anonymous · Answer

yw!