Question

Integration of 1/asin+bcosx

Answer 1

step 1:- asinx+bcosx=root(a^2+b^2){a/root (a^2+b^2) sinx+b/root (a^2+b^2)}
=root(a^2+b^2){sinx.cos@+cosx.sin@}
=root (a^2+b^2).sin(x+@)
where
cos@=a/root (a^2+b^2)
sin@=b/root (a^2+b^2)
hence @=tan^-1 (b/a)

step 2:-integrate 1/root (a^2+b^2) sin (x+@)
=1/root (a^2+b^2) integ(1/sin (x+@))
basic integration of cosec (x+@)=log|cosec (x+@)-cot(x+@)|
now integration of 1/asinx+bcosx

=1/root (a^2+b^2).log|cosec (x+@)-cot (x+@)|
now put @=tan^-1 (b/a)