Find the real positif number

**dhiab** · June 6th 2010, 07:25 PM

Find the real positif number a such that :
$<br /> \int_{0}^{a}\frac{tan(\frac{\pi }{4}+\frac{x}{2})}{sec^{2}(x)}dx=\frac{1}{16}<br />$

**Grandad** · June 6th 2010, 10:33 PM

Hello dhiab

Originally Posted by dhiab

Find the real positif number a such that :
$<br /> \int_{0}^{a}\frac{tan(\frac{\pi }{4}+\frac{x}{2})}{sec^{2}(x)}dx=\frac{1}{16}<br />$

There are many positive solutions. Here is the first one. But it doesn't work out very neatly, so check my working!

$\tan\left(\frac{\pi }{4}+\frac{x}{2}\right)=\frac{1+\tan\frac x2}{1-\tan\frac x2}$

$=\frac{\cos\frac x2+\sin\frac x2}{\cos\frac x2-\sin\frac x2}$

$=\frac{(\cos\frac x2+\sin\frac x2)^2}{\cos^2\frac x2-\sin^2\frac x2}$

$=\frac{1+2\sin\frac x2\cos\frac x2}{\cos x}$

$=\frac{1+\sin x}{\cos x}$

$\Rightarrow \int_0^a\frac{\tan\left(\frac{\pi }{4}+\frac{x}{2}\right)}{\sec^2x}\;dx = \int_0^a(\cos x + \cos x\sin x)\;dx$

$=\int_0^a(\cos x +\tfrac12 \sin2x)\;dx$

$=\Big[\sin x -\tfrac14\cos2x\Big]_0^a$

$=\sin a - \tfrac14\cos2a +\tfrac14$

$=\tfrac1{16}$

$\Rightarrow 16\sin a -4(1-2\sin^2a)+4=1$

$\Rightarrow 8\sin^2a+16\sin a -1=0$

$\Rightarrow \sin a = \frac{-16+\sqrt{256+32}}{16}$ , taking the positive root

$=\frac{-4+\sqrt{18}}{4}$

So the first positive solution is

$a = \arcsin\left(\frac{-4+\sqrt{18}}{4}\right)$

Grandad

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