#intarcsin(5x-2)dx#
Let #t=5x-2#
#dt=5dx#
#intarcsin(5x-2)dx=1/5int5arcsin(5x-2)dx#
#=1/5intarcsin(t)dt#
#=1/5int1*arcsin(t)dt#
Using Integration by parts :
#intg'(t)f(t)dt=[g(t)f(t)]-intg(t)f'(t)dt#
There:
#g'(t) =1# #f(t)=arcsin(t)#
#g(t)=t#, #f'(t)=1/sqrt(1-t²)#
So: #1/5intarcsin(t)dt=1/5([tarcsin(t)]-intt/sqrt(1-t²)dt)#
Now let #u=sqrt(1-t²)#
#du=-t/sqrt(1-t²)dt#
So:#1/5([tarcsin(t)]-intt/sqrt(1-t²)dt)=1/5([tarcsin(t)]+int1du)#
#=1/5([tarcsin(t)]+u)#
#=1/5(tarcsin(t)+sqrt(1-t²))#
#=1/5((5x-2)arcsin(5x-2)+sqrt(1-(5x-2)²))+C#, #C in RR#
\0/ here's our answer!
