# Thread: Find the value of Definite Integral

1. ## Find the value of Definite Integral

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

2. 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....

3. ## I am stuck

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 ???

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

5. ## Thank u all for helping

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

6. 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.

7. 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?????

8. 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.

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

10. Originally Posted by zorro
Find the value of
$
\int\limits_{\pi / 4}^{3 \pi / 4} \frac{1}{1+cos x}dx
$
$\int\frac{dx}{1+\cos(x)}=\int\frac{1-\cos(x)}{1-\cos^2(x)}dx=\int\left\{\frac{1}{\sin^2(x)}-\frac{\cos(x)}{\sin^2(x)}\right\}dx$

11. 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

12. 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.

13. 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).

14. 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 ....)

15. 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.