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Integration A Level

Integration of f'(x)/f (x) = ln f (x) + c

Integration of tanx would be ln cosx + c

how is integration of tan (x/2) = 2ln cos(x/2) why 2 as the coefficient of the term?
Original post by nonipaify
Integration of f'(x)/f (x) = ln f (x) + c

Integration of tanx would be ln cosx + c

how is integration of tan (x/2) = 2ln cos(x/2) why 2 as the coefficient of the term?


Integral of tanx is ln secx +c, or -ln cosx
Original post by nonipaify
Integration of f'(x)/f (x) = ln f (x) + c

Integration of tanx would be ln cosx + c

how is integration of tan (x/2) = 2ln cos(x/2) why 2 as the coefficient of the term?


Integrating tanx gives -ln cosx + C.

If you express tan(x/2) in terms of sin(x/2) and cos(x/2) it might help.
Reply 3
Avatar for davros
davros
16
Original post by nonipaify
Integration of f'(x)/f (x) = ln f (x) + c

Integration of tanx would be ln cosx + c

how is integration of tan (x/2) = 2ln cos(x/2) why 2 as the coefficient of the term?


Think about what happens when you differentiate ln[cos(x/2)]
If the integral of f(x) is g(x)+c then the integral of f(ax+b) is 1/a x g(ax+b) + c

The integral of tanx is -ln cosx + c

Also tan(x/2) is the same as tan(0.5x + 0)

The integral of tan(0.5x + 0) is therefore -ln cos(0.5x + 0) x 1/0.5 + c

As 1/0.5 is 2 and 0.5x is the same as x/2 this simplifies to -2ln cos(x/2) + c

I think the minus sign was accidentally missed out in the original question.
Reply 5
Avatar for ztibor
ztibor
10
Original post by nonipaify
Integration of f'(x)/f (x) = ln f (x) + c

Integration of tanx would be ln cosx + c

how is integration of tan (x/2) = 2ln cos(x/2) why 2 as the coefficient of the term?


∫tan⁡x2dx\int \tan \frac{x}{2} dx
means integration of a funtion with varable of x/2 w.r.t x ( dx is the integration factor showing you should to integrate with respect to x)
You can not integrate in this form only when the variable and the integration factor
are the same
1. We will integrate w.r.t x/2
then substitute the integration factor dx=d(x/2)*2 (from the d(x/2)/dx=1/2 differential)
so the original integral will be
2⋅∫tanx2d(x2)\displaystyle 2\cdot \int tan\frac{x}{2} d\left (\frac{x}{2}\right )

e.g when the intagration factor would be the tan function itself
∫tan⁡x2d(tan⁡x2)=tan⁡2x22+C\displaystyle \int \tan \frac{x}{2} d\left (\tan \frac{x}{2}\right )=\frac{\tan^2 \frac{x}{2}}{2}+C

2. substitute t=x/2 -> x=2t -> dx/dt=2 so dx will be 2dt
2⋅∫tantdt\displaystyle 2\cdot \int tan t dt

so the rule for
Integration of f(ax+b) w.r.t x will be

Unparseable latex formula:

\displaystyle \int f\left (ax+b\right ) dx = \frac{F\left (ax+b\rifgt)}{a}+C



for the question: so the divider of 1/2 will be the multplier of 2
(edited 9 years ago)
The numerous ways in which a problem such as this can be solved demonstrates the beauty of maths.
Reply 7
Avatar for nonipaify
nonipaify
OP
3
Thank you all for your replies my question has been cleared thanksssssss :smile:
Original post by Khallil
Mhm. Another way would be to use the Weierstrass substitution:

t=tan⁡x2  ⟹  dt=12sec⁡2x2 dx  ⟹  2dt1+t2=dxt = \tan \dfrac{x}{2} \implies \text{d}t = \dfrac{1}{2} \sec^2 \dfrac{x}{2} \text{ d}x \implies \dfrac{2\text{d}t}{1+t^2} = \text{d}x

tan⁡x=tan⁡(x2+x2)=2tan⁡x21−tan⁡2x2=2t1−t2\tan x = \tan \left( \dfrac{x}{2} + \dfrac{x}{2} \right) = \dfrac{2\tan \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} = \dfrac{2t}{1-t^2}

∴ ∫tan⁡x dx =t=tan⁡x2∫4t(1+t2)(1+t)(1−t) dt\displaystyle \begin{aligned} \therefore \ \int \tan x \text{ d}x \ & \overset{t = \tan \frac{x}{2}}= \int \dfrac{4t}{(1+t^2)(1+t)(1-t)} \text{ d}t \end{aligned}

Proceeding with a partial fraction decomposition will help to simplify the integrand quite nicely.

Edit: In case you're curious:

∴ ∫tan⁡x dx =t=tan⁡x2∫(2tt2+1−1t+1+11−t) dt   = log⁡∣t2+1∣−log⁡∣1−t∣−log⁡∣t+1∣+C   = log⁡∣t2+11−t2∣+C   = log⁡∣1+tan⁡2x21−tan⁡2x2∣+C   = log⁡∣1cos⁡2x2−sin⁡2x2∣+C   = log⁡∣1cos⁡x∣+C   = −log⁡∣cos⁡x∣+C\displaystyle \begin{aligned} \therefore \ \int \tan x \text{ d}x \ & \overset{t = \tan \frac{x}{2}}= \int \left( \dfrac{2t}{t^2 + 1} - \dfrac{1}{t+1} + \dfrac{1}{1-t} \right) \text{ d}t \\ & \ \ \ = \ \log | t^2 + 1 | - \log | 1-t | - \log | t+1 | + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{t^2 + 1}{1 - t^2} \right| + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{1 + \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} \right| + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{1}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} \right| + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{1}{\cos x} \right| + \mathcal{C} \\ & \ \ \ = \ - \log | \cos x | + \mathcal{C} \end{aligned}


where do you learn this cool stuff from?
Original post by Khallil
Mhm. Another way would be to use the Weierstrass substitution:

t=tan⁡x2  ⟹  dt=12sec⁡2x2 dx  ⟹  2dt1+t2=dxt = \tan \dfrac{x}{2} \implies \text{d}t = \dfrac{1}{2} \sec^2 \dfrac{x}{2} \text{ d}x \implies \dfrac{2\text{d}t}{1+t^2} = \text{d}x

tan⁡x=tan⁡(x2+x2)=2tan⁡x21−tan⁡2x2=2t1−t2\tan x = \tan \left( \dfrac{x}{2} + \dfrac{x}{2} \right) = \dfrac{2\tan \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} = \dfrac{2t}{1-t^2}

∴ ∫tan⁡x dx =t=tan⁡x2∫4t(1+t2)(1+t)(1−t) dt\displaystyle \begin{aligned} \therefore \ \int \tan x \text{ d}x \ & \overset{t = \tan \frac{x}{2}}= \int \dfrac{4t}{(1+t^2)(1+t)(1-t)} \text{ d}t \end{aligned}

Proceeding with a partial fraction decomposition will help to simplify the integrand quite nicely.

Edit: In case you're curious:

∴ ∫tan⁡x dx =t=tan⁡x2∫(2tt2+1−1t+1+11−t) dt   = log⁡∣t2+1∣−log⁡∣1−t∣−log⁡∣t+1∣+C   = log⁡∣t2+11−t2∣+C   = log⁡∣1+tan⁡2x21−tan⁡2x2∣+C   = log⁡∣1cos⁡2x2−sin⁡2x2∣+C   = log⁡∣1cos⁡x∣+C   = −log⁡∣cos⁡x∣+C\displaystyle \begin{aligned} \therefore \ \int \tan x \text{ d}x \ & \overset{t = \tan \frac{x}{2}}= \int \left( \dfrac{2t}{t^2 + 1} - \dfrac{1}{t+1} + \dfrac{1}{1-t} \right) \text{ d}t \\ & \ \ \ = \ \log | t^2 + 1 | - \log | 1-t | - \log | t+1 | + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{t^2 + 1}{1 - t^2} \right| + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{1 + \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} \right| + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{1}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} \right| + \mathcal{C} \\ & \ \ \ = \ \log \left| \dfrac{1}{\cos x} \right| + \mathcal{C} \\ & \ \ \ = \ - \log | \cos x | + \mathcal{C} \end{aligned}



That's really beautiful
Original post by Khallil
Mhm. Another way would be to use the Weierstrass substitution:


That's very nice, but I can't help but think of butterflies and wheels after reading that.
Original post by Khallil
What do you mean by butterflies and wheels? It's really elaborate, so the Weierstrass sub would usually be my last resort unless I was having a (really) bad day, haha!


It's from this poem. It means you could have achieved the same result more easily.
Original post by Khallil



I don't actually remember. I think I learnt it in the Mega A Level Maths Thread Mk III. It might have been Felix, joostan, Llewellyn, davros or another awesome person who taught it to me. I'll try and find the link to the explanation. It's pretty coherent.


I've always noticed you give an unusual (in a good way) method of doing a problem, like today, I wanna learn cool stuff like that too hence my username :smile:
Original post by Khallil
What do you mean by butterflies and wheels? It's really elaborate, so the Weierstrass sub would usually be my last resort unless I was having a (really) bad day, haha!


Well, Mr M has nicely provided the source of the expression. "breaking a butterfly on a wheel" mean to use vast overkill for a task. However, please don't take that as a criticism; it's very good practise to figure out as many different solutions as you can to a problem (though to be honest, once I've figured out one solution to a problem, I usually lose interest in doing it in other ways).

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