How to integrate these integrals

Question

$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$ $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$ It seems that substitutions make things worse:

$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$ $$ \Rightarrow \int \frac {-\sqrt{1 - (t-1)^2}}{t} = \int \frac {-\sqrt{t^2 + 2t}}{t} = \int \frac {-\sqrt t \cdot \sqrt t \cdot \sqrt{t + 2}}{\sqrt t \cdot t} $$

$$= \int \frac {- \sqrt{t + 2}}{\sqrt t } = \int - \sqrt{1 + \frac 2t} = ? $$

What next? I don’t know. Also, I’ve tried another “substitution”, namely $1 + \cos x = 2 \cos^2 \frac x2) $

$$ \int \frac {dx}{1+ \cos x} = \int \frac {dx}{2 \cos^2 \frac x2} = \int \frac 12 \cdot \sec^2 \frac x2 = ? $$ And failed again. Help me, please.

Answer 1

HINT:

$$\text{As }\frac{d(\tan mx)}{dx}=m\sec^2mx,$$

$$\int\sec^2mx= \frac{\tan mx}m+C$$

here $m=\frac12$

or using Weierstrass substitution $\tan \frac x2=t,$

$\frac x2=\arctan t\implies dx=2\frac{dt}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$

$$\int \frac{dx}{1+\cos x}=\int dt=t+K=\tan\frac x2+K$$

Answer 2

$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

Therefore, $$\int_0^{\pi/2}\dfrac{1}{1+\cos x}dx=\int_0^{\pi/2}\dfrac{1}{1+\sin x}dx$$

As you did $$\int_0^{\pi/2}\dfrac{1}{1+\cos x}dx=\frac{1}{2}\int_0^{\pi/2}\sec^2\frac{x}{2}dx$$

Substitute $x=2t$ and use $\int\sec^2 tdt=\tan t+C$

EDIT: Proof of $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$:

Let $I=\int_a^bf(x)dx$

Now, substitute $x=a+b-t\implies dx=-dt$

then $I=-\int_b^af(a+b-t)dt=\int_a^bf(a+b-t)dt=\int_a^bf(a+b-x)dx$$

Answer 3

$$ \begin{array}{lcl} \int_0^{\pi/2} \frac{1}{1+\cos x}dx &=& \int_0^{\pi/2} \frac{1}{(\cos^2(x/2)+\sin^2(x/2))+(\cos^2 (x/2)-\sin^2(x/2))}dx\\ &=& \int_0^{\pi/2} \frac{1}{2\cos^2(x/2)}dx\\ &=& \int_0^{\pi/4}\frac{1}{\cos^2u}du=\tan\frac{\pi}{4}=1 \end{array} $$