Indefinite Integrals
-
Integral of [ (1+cos^2(x)) / sin(x) ] dx
We know :-
sin^2(x) + cos^2(x)=1
So, cos^2(x)=1- sin^2(x)
So, our integral becomes:-
=Integral of ( (1 + 1 - sin^2(x))/sin(x))dx
=Integral of ( (2- sin^2(x))/sin(x))dx
=Integral of ( 2/sin(x) - sin(x))dx
=Integral of ( 2*cosec(x) - sin(x))dx
So, according to formulae:-
=2* log(cosec(x)+cot(x)) + cos(x)
We know :-
sin^2(x) + cos^2(x)=1
So, cos^2(x)=1- sin^2(x)
So, our integral becomes:-
=Integral of ( (1 + 1 - sin^2(x))/sin(x))dx
=Integral of ( (2- sin^2(x))/sin(x))dx
=Integral of ( 2/sin(x) - sin(x))dx
=Integral of ( 2*cosec(x) - sin(x))dx
So, according to formulae:-
=2* log(cosec(x)+cot(x)) + cos(x)
-
int [ 1 / sin x dx ] + int [ cos^2 x / sin x dx ]
= int [ 1 / sin x dx ] + int [ (1 - sin^2 x ) / sin x dx ]
= int [ 1 / sin x dx ] + int [ 1 / sin x dx ] - int [ sin x dx ]
= 2 * int [ csc x dx ] - int [ sin x dx ]
= ??
= int [ 1 / sin x dx ] + int [ (1 - sin^2 x ) / sin x dx ]
= int [ 1 / sin x dx ] + int [ 1 / sin x dx ] - int [ sin x dx ]
= 2 * int [ csc x dx ] - int [ sin x dx ]
= ??