How do I find the integral intcos(x)/(sin^2(x)+sin(x))dx∫cos(x)sin2(x)+sin(x)dx ?
1 Answer
=ln(sinx/(sin(x)+1))+c=ln(sinxsin(x)+1)+c , wherecc is a constantFull Answer
=intcos(x)/(sin^2(x)+sin(x))dx=∫cos(x)sin2(x)+sin(x)dx
=intcos(x)/(sin(x)(sin(x)+1))dx=∫cos(x)sin(x)(sin(x)+1)dx let's
sin(x)=tsin(x)=t , then,cos(x)dx=dtcos(x)dx=dt
=int1/(t(t+1))dt=∫1t(t+1)dt Using Partial Fractions,
1/(t(t+1))=A/t+B/(t+1)1t(t+1)=At+Bt+1 .............(i)(i) multiplying both sides with
t(t+1)t(t+1) ,
1=A(t+1)+Bt1=A(t+1)+Bt
1=A+(A+B)t1=A+(A+B)t comparing constant and coefficient on both sides, we get
A=1A=1 ,
A+B=0A+B=0 , which impliesB=-1B=−1 plugging the values of
AA andBB in(i)(i) ,
1/(t(t+1))=1/t-1/(t+1)1t(t+1)=1t−1t+1 integrating both sides with respect to
tt ,
int1/(t(t+1))dt=int1/tdt-int1/(t+1)dt∫1t(t+1)dt=∫1tdt−∫1t+1dt
=lnt-ln(t+1)+c=lnt−ln(t+1)+c , wherecc is a constant
=ln(t/(t+1))+c=ln(tt+1)+c , wherecc is a constantsubstituting
tt , yields
=ln(sinx/(sin(x)+1))+c=ln(sinxsin(x)+1)+c , wherecc is a constant
