How do you evaluate the integral #int 1/(sintheta+costheta)#?

2 Answers
May 1, 2018

#I=1/sqrt2ln|(tan(x/2)+sqrt2-1)/(tan(x/2)-sqrt2-1)|+c#

Explanation:

Here,

#I=int1/(sinx+cosx)dx#

Let,

#tan(x/2)=t=>sec^2(x/2)*1/2dx=dt#

#=>dx=(2dt)/sec^2(x/2)=(2dt)/(1+tan^2(x/2))=(2dt)/(1+t^2#

also, #sinx=(2tan(x/2))/(1+tan^2(x/2))=(2t)/(1+t^2)#

and #cosx=(1-tan^2(x/2))/(1+tan^2(x/2))=(1-t^2)/(1+t^2)#

So,

#I=int1/((2t)/(1+t^2)+(1-t^2)/(1+t^2))xx(2dt)/(1+t^2#

#=int2/(2t+1-t^2)dt#

#=2int1/(2-t^2+2t-1)dt#

#=2int1/((sqrt2)^2-(t-1)^2)dt#

#=2xx1/(2sqrt2)ln|(t-1+sqrt2)/(t-1-sqrt2)|+c#

Subst. back , #t =tan(x/2)#

#I=1/sqrt2ln|(tan(x/2)+sqrt2-1)/(tan(x/2)-sqrt2-1)|+c#

Note: For typing simplicity #x# is taken in place of #theta#.

May 1, 2018

#I=1/sqrt2ln|sec(theta-pi/4)+tan(theta-pi/4)|+c#

Explanation:

We know that,

#color(red)(cosC+cosD=2cos((C+D)/2)cos((C-D)/2)#

So,

#sintheta+costheta=color(red)(cos(pi/2-theta)+costheta#

#=color(red)(2cos((pi/2-theta+theta)/2)cos((pi/2-theta-theta)/2)#

#=2cos(pi/4)cos(pi/4-theta)#

#=2(1/sqrt2)cos(theta-pi/4)...to[as, cos(-alpha)=cosalpha ]#

#=sqrt2cos(theta-pi/4)#

Now,

#I=int1/(costheta+sintheta)d theta#

#=int1/(sqrt2cos(theta-pi/4))d theta#

#=1/sqrt2intsec(theta-pi/4)d theta#

#I=1/sqrt2ln|sec(theta-pi/4)+tan(theta-pi/4)|+c#

Note:

ans(1) and ans(2) are same but in different form.