Question

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

Answer 1

`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`.

Answer 2

`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.