# Thread: Today's calculation of integral #11

1. ## Today's calculation of integral #11

Let $\alpha>0,\,\beta<1$ and find $\int_{0}^{\pi/2}{2\sec x\cdot \ln \left[ \frac{1+\beta \cos x}{1+\alpha \cos x} \right]\,dx}.$

(We can kill this little problem with a simple trick.)

2. Hint (not actually a Hint): find an integral parameter.

3. I have a solution.

I don't know what an 'integral parameter' is, and the concept of differentiating inside the integral sign is fairly new to me, so bear with me.

Let $I_{n} = \int^{\frac{\pi}{2}}_{0} 2\sec x \ln(1 + n\cos x) \; \mathrm{d}x$

Then $\frac{\mathrm{d}I_{n}}{\mathrm{d}{n}} = \int^{\frac{\pi}{2}}_{0} \frac{2}{1 + n\cos x} \; \mathrm{d}x$

Using the substitution $t = \tan \frac{x}{2}$, this is equal to $4\left[\frac{1}{\sqrt{1-n^{2}}} \tan^{-1} \left( t\sqrt{\frac{1-n}{1+n}}\right) \right]^{1}_{0}$

$\Rightarrow I_{n} = \int \frac{4}{\sqrt{1-n^{2}}} \tan^{-1} \left(\sqrt{\frac{1-n}{1+n}}\right) \; \mathrm{d}n$

Noting that $\frac{\mathrm{d}}{\mathrm{d}n} \tan^{-1} \left(\sqrt{\frac{1-n}{1+n}}\right) = - \frac{1}{2\sqrt{1-n^{2}}}$, we have $I_{n} = c - 4\left(\tan^{-1}\sqrt{\frac{1-n}{1+n}}\right)^{2}$

$\Rightarrow I_{\beta} - I_{\alpha} = 4\left[\left(\tan^{-1}\sqrt{\frac{1-\alpha}{1+\alpha}}\right)^{2} - \left(\tan^{-1}\sqrt{\frac{1-\beta}{1+\beta}}\right)^{2} \right]$

This is equal to the integral that we have been asked to find.

When I get back at home I'll post another solution.

5. $\tan^{-1}\sqrt{\frac{1-x}{1+x}} = \cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{\cos^{-1}x}{2}$

So the answer can be expressed as

$\left(\cos^{-1}\alpha\right)^{2} - \left(\cos^{-1}\beta\right)^{2}$

Which is much nicer.

6. $2\sec x\cdot \ln \left[ \frac{1+\beta \cos x}{1+\alpha \cos x} \right]$

$2\sec x\cdot [\ln (1+\beta \cos x) - \ln (1+\alpha \cos x) ]$

$\int^{\beta}_{\alpha} \frac{2}{1+y\cos x}~dy$

So the problem becomes,

$\int^{\frac{\pi}{2}}_{0}\int^{\beta}_{\alpha} \frac{2}{1+y\cos x}~dy~dx$

Now change the integration order and so on..

7. Yes, that was the trick.