Find the real positif number

May 2009
596
31
ALGERIA
Find the real positif number a such that :
\(\displaystyle
\int_{0}^{a}\frac{tan(\frac{\pi }{4}+\frac{x}{2})}{sec^{2}(x)}dx=\frac{1}{16}
\)
 

Grandad

MHF Hall of Honor
Dec 2008
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South Coast of England
Hello dhiab
Find the real positif number a such that :
\(\displaystyle
\int_{0}^{a}\frac{tan(\frac{\pi }{4}+\frac{x}{2})}{sec^{2}(x)}dx=\frac{1}{16}
\)
There are many positive solutions. Here is the first one. But it doesn't work out very neatly, so check my working!

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

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


\(\displaystyle =\frac{1+2\sin\frac x2\cos\frac x2}{\cos x}\)


\(\displaystyle =\frac{1+\sin x}{\cos x}\)

\(\displaystyle \Rightarrow \int_0^a\frac{\tan\left(\frac{\pi }{4}+\frac{x}{2}\right)}{\sec^2x}\;dx = \int_0^a(\cos x + \cos x\sin x)\;dx\)
\(\displaystyle =\int_0^a(\cos x +\tfrac12 \sin2x)\;dx\)

\(\displaystyle =\Big[\sin x -\tfrac14\cos2x\Big]_0^a\)


\(\displaystyle =\sin a - \tfrac14\cos2a +\tfrac14\)


\(\displaystyle =\tfrac1{16}\)

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

\(\displaystyle \Rightarrow 8\sin^2a+16\sin a -1=0\)


\(\displaystyle \Rightarrow \sin a = \frac{-16+\sqrt{256+32}}{16}\), taking the positive root
\(\displaystyle =\frac{-4+\sqrt{18}}{4}\)
So the first positive solution is
\(\displaystyle a = \arcsin\left(\frac{-4+\sqrt{18}}{4}\right)\)
Grandad