Find the value of
$\displaystyle
\int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx
$
$\displaystyle \int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}}$ $\displaystyle \frac{1}{1 + \cos x} \, dx $
since $\displaystyle \cos x = \cos \left(2 \cdot \frac{x}{2}\right)$
$\displaystyle = \frac{1}{1 + \cos \left(2 \cdot \frac{x}{2}\right)}$
since $\displaystyle \cos(2A) = 2 \cos^2 A - 1$
$\displaystyle = $ $\displaystyle \int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}}\frac{1}{1 + 2 \cos^2 \left(\frac{x}{2}\right) - 1} \, dx$
= $\displaystyle \int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}} \frac{1}{2} \sec^2 \left( \frac{x}{2}\right) \, dx$
= $\displaystyle \frac{1}{2} \left[ \tan \frac{3 \pi}{8} - \tan \frac{\pi}{8} \right]$........I am stuck here ???
First we can do the indefinite integral. Let $\displaystyle t=\tan{\frac x 2}$, so $\displaystyle dt = \frac{1}{2}(1+t^2)dx$. We have also $\displaystyle \cos x = \frac{1-t^2}{1+t^2}$.
$\displaystyle \int \frac{dx}{1+cos x} = \int\frac{2dt}{(1+t^2)}\times\frac{1}{1+\frac{1-t^2}{1+t^2}} $
$\displaystyle = \int \frac{2dt}{(1+t^2)+(1-t^2)} = \int dt = t + C = \tan{\frac x 2}+C$.
So $\displaystyle \int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx = \tan{\frac{3\pi}{8}}-\tan{\frac{\pi}{8}}$
The blue 1/2 should not be there. $\displaystyle \int \sec^2 \left( \frac{x}{2} \right) \, dx = \tan \left( \frac{x}{2} \right)$ NOT $\displaystyle \frac{1}{2} \tan \left( \frac{x}{2} \right)$.
By the way, the double angle formula can be used to get an exact surd value of $\displaystyle \tan \frac{\pi}{8}$:
$\displaystyle \tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}}$.
So solve $\displaystyle 1 = \frac{2x}{1 - x^2}$ for x and discard the negative solution.
I am getting this answer
$\displaystyle \frac{1}{2} \left[ tan \left(\frac{3 \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$ =
$\displaystyle \frac{1}{2} \left[ tan \left(\frac{\pi}{2} + \frac{ \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$ =
$\displaystyle \frac{1}{2} \left[ - cot \left( \frac{ \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$=
$\displaystyle \frac{1}{2} \left[ - 1 - 1 \right]$= $\displaystyle -1$ .................. Is this correct?????
Hello,
First substitute $\displaystyle t=x-\tfrac \pi 2$
Then $\displaystyle \cos(x)=-\sin(t)$
So the integral I becomes :
$\displaystyle I=\int_{-\frac\pi4}^{\frac\pi4} \frac{1}{1-\sin(t)} ~dt$
Similarly, if we substitute $\displaystyle t=\tfrac\pi 2-x$, we get $\displaystyle I=\int_{-\frac\pi4}^{\frac\pi4} \frac{1}{1+\sin(t)} ~dt$
So $\displaystyle 2I=\int_{-\frac\pi4}^{\frac\pi4} \frac{1}{1+\sin(t)}+ \frac{1}{1-\sin(t)} ~dt=\int_{-\frac\pi4}^{\frac\pi4} \frac{2 ~dt}{1-\sin^2(t)}$
Hence $\displaystyle I=\int_{-\frac\pi4}^{\frac\pi4} \frac{dt}{\cos^2(t)}=\left.\tan(t)\right|_{-\frac\pi4}^{\frac\pi4}=\boxed{2}$
I hope it's not too wrong... It's been a very long time since I've done any of these
But at least the result looks more beautiful than those above
No. From the usual double angle formula it's $\displaystyle \tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8} }{1 - \tan^2 \frac{\pi}{8}}$ from which it follows that $\displaystyle \tan \frac{\pi}{8} = \sqrt{2} - 1$.
And since $\displaystyle \tan \frac{3 \pi}{8} = \tan \left( \frac{\pi}{4} + \frac{\pi}{8}\right)$ it follows from the compound angle formula that $\displaystyle \tan \frac{3 \pi}{8} = \sqrt{2} + 1$.
From which the correct answer of 2 follows. You are going to continue to struggle with these 'higher level' questions if you don't start extensively reviewing the 'lower level' material that underpins these questions.
I also agree. Totally. But I wanted to validate the method used by the OP. (The Weierstrass substitution is the routine cookbook way that students are taught to do these - using conjugates might, to the OP, look too much like pulling rabbits out of a hat ....)