$\int$ (sin x+ tan x)^-1dx

gokusuper

studyid

simple sum ,

Given integral = $\int\frac{cosx}{sinx(1+cosx)}dx\:\:\:or\:\int\frac{dx}{tanx(1+cosx)}$

Simply put $tan(\frac{x}{2})=t \:\:\:and \:proceed.$

chinmay_saxena01

hey can u explain further???????

devil_angel

arey rite cos x in terms of tan x/2 .nd relace by it nd we know differenciation of tanx/2 will come in terms of sec²(x/2)

wich is 1 + tan²(x/2)

chinmay_saxena01

ok thanks

studyid

$\int\frac{dx}{tanx(1+cosx)}$

$Putting\:tan(\frac{x}{2})=t\:we\:have\:dx=\frac{2dt}{1+{t}^{2}}$

$Also\:cosx=\frac{1-tan^{2}({\frac{x}{2}})}{1+tan^{2}({\frac{x}{2}})}=\frac{1-t^{2}}{1+t^{2}}$

$And\:tanx=\frac{2tan{\frac{x}{2}}}{1-tan^{2}(\frac{x}{2})}=\frac{2t}{1-t^{2}}$

$Hence\:given\:integral\:turns\:into\:$ :

$\int\frac{\frac{2dt}{1+t^{2}}}{\frac{2t}{1-t^{2}}(1+\frac{1-t^{2}}{1+t^{2}})}$

$Taking\:lcm\:and\:simplifying\:we\:get\:$

$\frac{1}{2}\int\frac{1-t^{2}}{t}dt\\which\:is\:a\:simple\:integral.$

(sin x+ tan x)-1 dx