How can I integrate (1-sin x)/cosx?

Question

Steps Please!

Answer 1

♠ y(x)*dx = dx*(1-sin x)/cosx =dx/cosx –dx*sinx/cosx =y1*dx+y2*dx;
y1*dx= dx/cosx;
t=tan(x/2), dt=0.5dx/ (cos(x/2))^2 =0.5dx*(1+tt),
hence dx=2dt/(1+tt), cosx =(1-tt)/(1+tt); thus
♣ y1(t)*dt =2dt/(1-tt) =dt/(1-t) +dt/(1+t), hence
Y(t) = ln|(1+t)/(1-t)| =ln|(1+t)^2/(1-t^2)|;
Y(x) = ln|((cos(x/2))^2 +2cos(x/2)*sin(x/2) +(sin(x/2))^2)/
/((cos(x/2))^2 –(sin(x/2))^2) = ln|(1+sinx)/cosx|;
♦ y2(x)*dx =-dx*sinx/cosx =d(cosx)/cosx =d(ln|cosx|);
♥ Y(x) = ln|(1+sinx)/cosx| +ln|cosx| =ln|1+sinx| +C;
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Answer 2

∫(1-sin x)/cosx dx

distribute the 1/cos(x) to seperate it

∫1/cos(x) dx - ∫sin(x)/cos(x) dx

Simplify slightly
∫sec(x)dx - ∫tan(x)dx

Integrate
ln |sec(x) + tan(x)| - ln | sec(x) | + C

Answer 3

Try breaking it up into two, 1/cos x and (sin x)/(cos x). And take it from there.