# How to integrate these integrals

$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$ $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$ 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.

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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 $\cos x=\frac{1-t^2}{1+t^2}$

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

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Thanks. Let's see: $\Rightarrow \tan (x\frac 12) \cdot 2 = 2 (\tan \frac {\pi} {4} - \tan 0) = 2$ –  epsigma Jul 2 '13 at 5:42
@epsigma, also as $t=\tan\frac x2, x=0\implies t=0$ and $x=\frac\pi2\implies t=1$ So, $\int_0^1dt=t\big|_0^1=1-0=1$ –  lab bhattacharjee Jul 2 '13 at 6:05
It should be $$\int\frac{1}{1+\cos x}\;dx=\tan\frac{x}{2}+K.$$. Indeed, you're correct about the substitution except for the last part. –  Tunk-Fey Mar 17 '14 at 6:06
@Tunk-Fey,thanks for your observation –  lab bhattacharjee Mar 17 '14 at 6:19

You were on the right track in the last line

$$\frac{1}{2} \int \sec^2 \frac{x}{2} dx = \tan \frac{x}{2} + C$$

That's it. Put the limits in and you're done.

For the second one, substitute $t = \frac{\pi}{2} - x$ and you're back to the first integral

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Use this simple trigonometry manipulation: $$\frac{1}{1+\cos x}=\frac{1}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}=\frac{1-\cos x}{\sin^2 x}.$$ Therefore \begin{align} \int\frac{1}{1+\cos x}\;dx&=\int\frac{1-\cos x}{\sin^2 x}\;dx\\ &=\int\frac{1}{\sin^2 x}\;dx-\int\frac{\cos x}{\sin^2 x}\;dx\\ &=-\int\;d(\cot x)-\int\frac{1}{\sin^2 x}\;d(\sin x)\\ &=-\cot x+\frac{1}{\sin x}+C\\ &=\frac{1-\cos x}{\sin x}+C\\ &=\frac{\sin^2\frac{1}{2}x+\cos^2\frac{1}{2}x-\cos^2\frac{1}{2}x+\sin^2\frac{1}{2}x}{2\sin\frac{1}{2}x\cos\frac{1}{2}x}+C\\ &=\tan\frac{1}{2}x+C. \end{align} Similarly with $$\frac{1}{1+\sin x}=\frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{\cos^2 x}.$$

$$\text{# }\mathbb{Q.E.D.}\text{ #}$$

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

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

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$$- How do you knowthis:$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$– epsigma Jul 2 '13 at 5:47 Let$u = a + b - x\$ –  Hawk Jul 2 '13 at 5:58