# intergral help?

integrate

{cos^2 (x)}/(1+tanx)

plz show the working

Relevance
• Pascal
Lv 7

∫cos² x/(1+tan x) dx

1/2 ∫2 cos² x/(1+tan x) dx

1/2 ∫(cos (2x)+1)/(1+tan x) dx

1/2 ∫cos x (cos (2x) + 1)/(cos x + sin x) dx

1/2 ∫cos x (cos (2x) + 1)(cos x - sin x)/(cos² x - sin² x) dx

1/2 ∫cos x (cos (2x) + 1)(cos x - sin x)/cos (2x) dx

1/2 ∫cos x (cos x - sin x) + cos x (cos x - sin x)/cos (2x) dx

1/2 ∫cos² x - cos x sin x + cos x (cos x - sin x)/cos (2x) dx

1/2 ∫cos² x - cos x sin x + (cos² x - cos x sin x)/cos (2x) dx

1/2 ∫cos² x - cos x sin x + (cos (2x)/2 + 1/2 - cos x sin x)/cos (2x) dx

1/2 ∫cos² x - cos x sin x + 1/2 + (1/2 - cos x sin x)/cos (2x) dx

1/2 ∫cos² x - cos x sin x + 1/2 + 1/(2 cos (2x)) - cos x sin x/cos (2x) dx

1/2 ∫cos² x - cos x sin x + 1/2 + 1/(2 cos (2x)) - sin (2x)/(2 cos (2x)) dx

1/2 ∫cos² x - cos x sin x + 1/2 + 1/2 sec (2x) - 1/2 tan (2x) dx

1/2 ∫cos (2x)/2 - cos x sin x + 1 + 1/2 sec (2x) - 1/2 tan (2x) dx

1/2 (sin (2x)/4 + cos² x/2 + x + ln |sec (2x) + tan (2x)|/4 + ln |cos (2x)/4) + C

sin (2x)/8 + cos² x/4 + x/2 + ln |sec (2x) + tan (2x)|/8 + ln |cos (2x)|/8 + C

Substitute u=e^(ix) so then du=ie^(ix)dx and dx = 1/(iu)*du and cos(x)^2/(1+tan(x)) = ((u+1/u)/2)^2 /(1+(u-1/u)/(u+1/u)) = ((u^2+1)/(2u))^2 / (2u^2/(u^2+1)) = (u^2+1)^3 /(8u^3) = (u^6+3u^4 +3u^2+1)/ (8u^3) so then you integrate 1/(iu)*that which is (1/i) * (1/8u^2 +3/8 +3/8u^(-2) +1/8u^(-4)0 then use the power rule to get (1/i) *(1/24u^3 +3/8u -3/8u^-1 - 1/24u^-3) oh shoot there is an i term from the sine function ugggg so you start at the beginning again with the i term and fix it. My recommendation is to use technology for these type integrals.

• Anonymous

So y(x)=0.5(1 +cos2x) cosx / (cosx +sinx); cosx +sinx = sqrt2 sin(pi/4+x);

Let t=pi/4+x, then x=t - pi/4, dx=dt;

So y(t) = (0.5/sqrt2) (1+ cos(pi/2-2t)) cos(t-pi/4) / sint =

= (0.5/sqrt2) (1+ sin(2t)) (cost/sqrt2 +sint/sqrt2) / sint =

= 0.25 (1+ sin(2t)) (cost +sint) / sint = 0.25{cotan(t) +1 +2(cost)^2 +sin(2t)} =

= 0.25{cotan(t) +1 +cos(2t)+1 +sin(2t)} =

= 0.25{cotan(t) +cos(2t) +sin(2t) +2};

Y(t) =0.25{ln|sint| +0.5sin(2t) -0.5cos(2t) +2t} +C;

Y(t) =0.25{ln|sin(x+pi/4)| +0.5cos(2x) +0.5sin(2x) +2x+pi/2} +C;

Y(t) =0.25ln|sin(x+pi/4)| +0.125(cos(2x) +sin(2x)) +0.5x +C1;