Find the value of Definite Integral

**zorro** · May 3rd 2009, 10:39 PM

Find the value of
$ \int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx $

**stapel** · May 4th 2009, 04:17 AM

Convert cos(x) to cos(2(x/2)) and substitute using a double-angle identity to rearrange the denominator to get 2cos^2(x/2). Then the fraction can be restated as (1/2)sec^2(x/2).

Then apply what you know about the derivative of the tangent....

**zorro** · December 20th 2009, 12:24 PM

Originally Posted by stapel

Convert cos(x) to cos(2(x/2)) and substitute using a double-angle identity to rearrange the denominator to get 2cos^2(x/2). Then the fraction can be restated as (1/2)sec^2(x/2).

Then apply what you know about the derivative of the tangent....

$\int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}}$ $\frac{1}{1 + \cos x} \, dx$

since $\cos x = \cos \left(2 \cdot \frac{x}{2}\right)$

$= \frac{1}{1 + \cos \left(2 \cdot \frac{x}{2}\right)}$

since $\cos(2A) = 2 \cos^2 A - 1$

$=$ $\int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}}\frac{1}{1 + 2 \cos^2 \left(\frac{x}{2}\right) - 1} \, dx$

= $\int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}} \frac{1}{2} \sec^2 \left( \frac{x}{2}\right) \, dx$

= $\frac{1}{2} \left[ \tan \frac{3 \pi}{8} - \tan \frac{\pi}{8} \right]$ ........I am stuck here ???

**Bruno J.** · December 20th 2009, 01:44 PM

Originally Posted by zorro

Find the value of
$ \int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx $

First we can do the indefinite integral. Let $t=\tan{\frac x 2}$ , so $dt = \frac{1}{2}(1+t^2)dx$ . We have also $\cos x = \frac{1-t^2}{1+t^2}$ .

$\int \frac{dx}{1+cos x} = \int\frac{2dt}{(1+t^2)}\times\frac{1}{1+\frac{1-t^2}{1+t^2}}$

$= \int \frac{2dt}{(1+t^2)+(1-t^2)} = \int dt = t + C = \tan{\frac x 2}+C$ .

So $\int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx = \tan{\frac{3\pi}{8}}-\tan{\frac{\pi}{8}}$

**zorro** · December 20th 2009, 02:19 PM

Originally Posted by Bruno J.

First we can do the indefinite integral. Let $t=\tan{\frac x 2}$ , so $dt = \frac{1}{2}(1+t^2)dx$ . We have also $\cos x = \frac{1-t^2}{1+t^2}$ .

$\int \frac{dx}{1+cos x} = \int\frac{2dt}{(1+t^2)}\times\frac{1}{1+\frac{1-t^2}{1+t^2}}$

$= \int \frac{2dt}{(1+t^2)+(1-t^2)} = \int dt = t + C = \tan{\frac x 2}+C$ .

So $\int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx = \tan{\frac{3\pi}{8}}-\tan{\frac{\pi}{8}}$

thank u all for helping
cheers

**mr fantastic** · December 21st 2009, 03:35 AM

Originally Posted by zorro

$\int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}}$ $\frac{1}{1 + \cos x} \, dx$

since $\cos x = \cos \left(2 \cdot \frac{x}{2}\right)$

$= \frac{1}{1 + \cos \left(2 \cdot \frac{x}{2}\right)}$

since $\cos(2A) = 2 \cos^2 A - 1$

$=$ $\int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}}\frac{1}{1 + 2 \cos^2 \left(\frac{x}{2}\right) - 1} \, dx$

= $\int^{\frac{3 \pi}{4}}_{\frac{\pi}{4}} \frac{1}{2} \sec^2 \left( \frac{x}{2}\right) \, dx$

= ${\color{blue}\frac{1}{2}} \left[ \tan \frac{3 \pi}{8} - \tan \frac{\pi}{8} \right]$ ........I am stuck here ???

The blue 1/2 should not be there. $\int \sec^2 \left( \frac{x}{2} \right) \, dx = \tan \left( \frac{x}{2} \right)$ NOT $\frac{1}{2} \tan \left( \frac{x}{2} \right)$ .

By the way, the double angle formula can be used to get an exact surd value of $\tan \frac{\pi}{8}$ :

$\tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}}$ .

So solve $1 = \frac{2x}{1 - x^2}$ for x and discard the negative solution.

**zorro** · December 27th 2009, 12:09 PM

I am getting this answer

$\frac{1}{2} \left[ tan \left(\frac{3 \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$ =

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

$\frac{1}{2} \left[ - cot \left( \frac{ \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$ =

$\frac{1}{2} \left[ - 1 - 1 \right]$ = $-1$ .................. Is this correct?????

**mr fantastic** · December 27th 2009, 02:44 PM

Originally Posted by zorro

I am getting this answer

$\frac{1}{2} \left[ tan \left(\frac{3 \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$ =

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

$\frac{1}{2} \left[ - cot \left( \frac{ \pi}{4} \right) - tan \left( \frac{\pi}{4} \right) \right]$ =

$\frac{1}{2} \left[ - 1 - 1 \right]$ = $-1$ .................. Is this correct?????

The correct nswer is $\tan \frac{3 \pi}{8} - \tan \frac{\pi}{8}$ and I have explained how to get the exact surd values of each term, should you wish to give an answer in an exact surd form. I don't know where you got the 3pi/4 and pi/4 from.

**zorro** · December 27th 2009, 09:19 PM

Originally Posted by mr fantastic

The blue 1/2 should not be there. $\int \sec^2 \left( \frac{x}{2} \right) \, dx = \tan \left( \frac{x}{2} \right)$ NOT $\frac{1}{2} \tan \left( \frac{x}{2} \right)$ .

By the way, the double angle formula can be used to get an exact surd value of $\tan \frac{\pi}{8}$ :

$\tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}}$ .

So solve $1 = \frac{2x}{1 - x^2}$ for x and discard the negative solution.

In the above shouldnt it be $tan \frac{\pi}{8} = \frac{2 tan \frac{\pi}{8} }{1 - tan^2 \frac{\pi}{8}}$

**Drexel28** · December 27th 2009, 11:34 PM

Originally Posted by zorro

Find the value of
$ \int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx $

$\int\frac{dx}{1+\cos(x)}=\int\frac{1-\cos(x)}{1-\cos^2(x)}dx=\int\left\{\frac{1}{\sin^2(x)}-\frac{\cos(x)}{\sin^2(x)}\right\}dx$

**Moo** · December 28th 2009, 05:03 AM

Hello,

First substitute $t=x-\tfrac \pi 2$

Then $\cos(x)=-\sin(t)$

So the integral I becomes :

$I=\int_{-\frac\pi4}^{\frac\pi4} \frac{1}{1-\sin(t)} ~dt$

Similarly, if we substitute $t=\tfrac\pi 2-x$ , we get $I=\int_{-\frac\pi4}^{\frac\pi4} \frac{1}{1+\sin(t)} ~dt$

So $2I=\int_{-\frac\pi4}^{\frac\pi4} \frac{1}{1+\sin(t)}+ \frac{1}{1-\sin(t)} ~dt=\int_{-\frac\pi4}^{\frac\pi4} \frac{2 ~dt}{1-\sin^2(t)}$

Hence $I=\int_{-\frac\pi4}^{\frac\pi4} \frac{dt}{\cos^2(t)}=\left.\tan(t)\right|_{-\frac\pi4}^{\frac\pi4}=\boxed{2}$

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

**matheagle** · December 28th 2009, 10:04 PM

Originally Posted by Drexel28

$\int\frac{dx}{1+\cos(x)}=\int\frac{1-\cos(x)}{1-\cos^2(x)}dx=\int\left\{\frac{1}{\sin^2(x)}-\frac{\cos(x)}{\sin^2(x)}\right\}dx$

I agree, I think the way to do this is with conjugates.

**Drexel28** · December 28th 2009, 10:05 PM

Originally Posted by matheagle

I agree, I think the way to do this is with conjugates.

I agree, for the indefinite case. But unlike me Moo noticed that this was a definite integral and thus subject to all of the nice little tricks that go along with it (symmetry in this case).

**mr fantastic** · December 28th 2009, 10:07 PM

Originally Posted by zorro

In the above shouldnt it be $tan \frac{\pi}{8} = \frac{2 tan \frac{\pi}{8} }{1 - tan^2 \frac{\pi}{8}}$

No. From the usual double angle formula it's $\tan \frac{\pi}{4} = \frac{2 \tan \frac{\pi}{8} }{1 - \tan^2 \frac{\pi}{8}}$ from which it follows that $\tan \frac{\pi}{8} = \sqrt{2} - 1$ .

And since $\tan \frac{3 \pi}{8} = \tan \left( \frac{\pi}{4} + \frac{\pi}{8}\right)$ it follows from the compound angle formula that $\tan \frac{3 \pi}{8} = \sqrt{2} + 1$ .

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.

Originally Posted by matheagle

I agree, I think the way to do this is with conjugates.

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 ....)

**matheagle** · December 28th 2009, 10:17 PM

I like having a bag of tricks like that.
I reviewed Salas, Hille last year and I pointed out that they omitted that technique.
I don't think they included it even after their revisions.
They should have mentioned it.

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Find the value of Definite Integral

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