+105 I have to integrate the inverse sine of x but I have to do it by parts ???
How do I do that?
+118703 the formula is
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dxor∫uv′dx=uv−∫u′vdx
∫sin−1dx=∫sin−1∗1dxletu=sin−1andv′=1u=sin−1xx=sinudxdu=cosududx=1cosududx=1√cos2ududx=1√1−sin2ududx=1√1−x2
sou=sin−1xu′=1√1−x2v′=1v=x∫sin−1xdx=∫sin−1∗1dx=uv−∫u′vdx=sin−1x∗x−∫x√1−x2dx=xsin−1x−∫x√1−x2dxlett=1−x2dtdx=−2xdx=dt−2x=dt−2√1−t√1−t=x
soxsin−1x−∫x√1−x2dx=xsin−1x−∫√1−t√tdt−2√1−t=xsin−1x−∫√1−t√t(−2√1−t)dt=xsin−1x−∫√1−t−2√t√1−tdt=xsin−1x−∫1−2√tdt=xsin−1x+12∫t−0.5dt=xsin−1x+12×t0.50.5+c=xsin−1x+√t+c=xsin−1x+√1−x2+c
+118703 the formula is
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dxor∫uv′dx=uv−∫u′vdx
∫sin−1dx=∫sin−1∗1dxletu=sin−1andv′=1u=sin−1xx=sinudxdu=cosududx=1cosududx=1√cos2ududx=1√1−sin2ududx=1√1−x2
sou=sin−1xu′=1√1−x2v′=1v=x∫sin−1xdx=∫sin−1∗1dx=uv−∫u′vdx=sin−1x∗x−∫x√1−x2dx=xsin−1x−∫x√1−x2dxlett=1−x2dtdx=−2xdx=dt−2x=dt−2√1−t√1−t=x
soxsin−1x−∫x√1−x2dx=xsin−1x−∫√1−t√tdt−2√1−t=xsin−1x−∫√1−t√t(−2√1−t)dt=xsin−1x−∫√1−t−2√t√1−tdt=xsin−1x−∫1−2√tdt=xsin−1x+12∫t−0.5dt=xsin−1x+12×t0.50.5+c=xsin−1x+√t+c=xsin−1x+√1−x2+c
