Question

integration of {(sinx -xcosx)/x(x+sinx)}

Answer 1

hope this hlp............................

Answer 2

Integration of {(sinx -xcosx)/x(x+sinx)} is sinx+C

Step-by-step explanation:

∫$\frac{sinx+x-x-xcosx}{x(x+sinx)}dx$

=∫$\frac{sinx+x-x-xcosx}{x(x+sinx)}dx$

=∫$\frac{(sinx+x)- (x+xcosx)}{x(x+sinx)}dx$

=∫$\frac{(sinx+x)}{x(x+sinx)} -\frac{(x+xcosx)}{x(x+sinx)}dx$

=∫$\frac{1}{x}-\frac{(x+xcosx)}{x(x+sinx)}dx$

=∫$\frac{1}{x}dx$ -∫$\frac{(x+xcosx)}{x(x+sinx)}dx$

=∫$\frac{1}{x}dx$ -∫$\frac{x(1+cosx)}{x(x+sinx)}dx$

Cancelling x in second integration.

=$\ln|x|$-∫$\frac{1+cosx}{x+sinx} dx$

Taking x+sinx=t

(1+cosx) dx=dt

Therefore $\ln|x|$-∫$\frac{1}{t} dt$

=$\ln|x-\ln|t|+C$

Now putting t=x+sinx

=ln|x| - ln|x|+sinx+C

=sinx+C