How do you evaluate the integral $\int \frac{1}{sin θ + cos θ}$ $int 1/(sintheta+costheta)$ ?

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How do you evaluate the integral $\int \frac{1}{sin θ + cos θ}$ $int 1/(sintheta+costheta)$ ?

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Answer 1 · 2018-05-01T10:47:59

$I = \frac{1}{\sqrt{2}} ln ∣ ∣ ∣ ∣ \frac{tan (\frac{x}{2}) + \sqrt{2} - 1}{tan (\frac{x}{2}) - \sqrt{2} - 1} ∣ ∣ ∣ ∣ + c$ $I=1/sqrt2ln|(tan(x/2)+sqrt2-1)/(tan(x/2)-sqrt2-1)|+c$

Explanation:

Here,

$I = \int \frac{1}{sin x + cos x} d x$ $I=int1/(sinx+cosx)dx$

Let,

$tan (\frac{x}{2}) = t \Rightarrow {sec}^{2} (\frac{x}{2}) \cdot \frac{1}{2} d x = d t$ $tan(x/2)=t=>sec^2(x/2)*1/2dx=dt$

$\Rightarrow d x = \frac{2 d t}{{sec}^{2} (\frac{x}{2})} = \frac{2 d t}{1 + {tan}^{2} (\frac{x}{2})} = \frac{2 d t}{1 + t^{2}}$ $=>dx=(2dt)/sec^2(x/2)=(2dt)/(1+tan^2(x/2))=(2dt)/(1+t^2$

also, $sin x = \frac{2 tan (\frac{x}{2})}{1 + {tan}^{2} (\frac{x}{2})} = \frac{2 t}{1 + t^{2}}$ $sinx=(2tan(x/2))/(1+tan^2(x/2))=(2t)/(1+t^2)$

and $cos x = \frac{1 - {tan}^{2} (\frac{x}{2})}{1 + {tan}^{2} (\frac{x}{2})} = \frac{1 - t^{2}}{1 + t^{2}}$ $cosx=(1-tan^2(x/2))/(1+tan^2(x/2))=(1-t^2)/(1+t^2)$

So,

$I = \int \frac{1}{\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}}$ $I=int1/((2t)/(1+t^2)+(1-t^2)/(1+t^2))xx(2dt)/(1+t^2$

$= \int \frac{2}{2 t + 1 - t^{2}} d t$ $=int2/(2t+1-t^2)dt$

$= 2 \int \frac{1}{2 - t^{2} + 2 t - 1} d t$ $=2int1/(2-t^2+2t-1)dt$

$= 2 \int \frac{1}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} d t$ $=2int1/((sqrt2)^2-(t-1)^2)dt$

$= 2 \times \frac{1}{2 \sqrt{2}} ln ∣ ∣ ∣ \frac{t - 1 + \sqrt{2}}{t - 1 - \sqrt{2}} ∣ ∣ ∣ + c$ $=2xx1/(2sqrt2)ln|(t-1+sqrt2)/(t-1-sqrt2)|+c$

Subst. back , $t = tan (\frac{x}{2})$ $t =tan(x/2)$

$I = \frac{1}{\sqrt{2}} ln ∣ ∣ ∣ ∣ \frac{tan (\frac{x}{2}) + \sqrt{2} - 1}{tan (\frac{x}{2}) - \sqrt{2} - 1} ∣ ∣ ∣ ∣ + c$ $I=1/sqrt2ln|(tan(x/2)+sqrt2-1)/(tan(x/2)-sqrt2-1)|+c$

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

Answer 2 · 2018-05-01T11:15:17

$I = \frac{1}{\sqrt{2}} ln ∣ ∣ sec (θ - \frac{π}{4}) + tan (θ - \frac{π}{4}) ∣ ∣ + c$ $I=1/sqrt2ln|sec(theta-pi/4)+tan(theta-pi/4)|+c$

Explanation:

We know that,

$cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2})$ $color(red)(cosC+cosD=2cos((C+D)/2)cos((C-D)/2)$

So,

$sin θ + cos θ = cos (\frac{π}{2} - θ) + cos θ$ $sintheta+costheta=color(red)(cos(pi/2-theta)+costheta$

$= 2 cos (\frac{\frac{π}{2} - θ + θ}{2}) cos (\frac{\frac{π}{2} - θ - θ}{2})$ $=color(red)(2cos((pi/2-theta+theta)/2)cos((pi/2-theta-theta)/2)$

$= 2 cos (\frac{π}{4}) cos (\frac{π}{4} - θ)$ $=2cos(pi/4)cos(pi/4-theta)$

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

$=sqrt2cos(theta-pi/4)$

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.

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