∫sin−1x(1−x2)3/4dx
Let I=∫sin−1x(1−x2)3/4dx=∫sin−1x(1−x2)√1−x2dxPut sin−1x=t⇒1√1−x2dx=dtand x=sin t⇒1−x2=cos2t⇒cost=√1−x2∴ I=∫tcos2tdt
Using integration by parts,
I=t∫sec2t dt−∫(ddtt.∫sec2tdt)dt=t.tant−∫tan tdt=t.tant−∫1.tan t dt=t.tant+log|cos t|+C [∵∫tan x dx=−log|cosx|+C]=sin−1x.x√1−x2+log|√1−x2|C