$\displaystyle \int 1/(7-5sin(x))\ dx$
What is method of this? Many helps
Hi,
Indeed this is a tricky integral that can be solved via substitution.
Steps
- $\displaystyle \int{\dfrac{1}{7-5\sin(x)}}dx$
- Substitution
$\displaystyle u=\tan\dfrac{x}{2}$.
Then, $\displaystyle du=\dfrac{1}{2}\sec^2(\dfrac{x}{2})dx$
And, $\displaystyle \sin(x)=\dfrac{2u}{u^2+1}, dx=\dfrac{2du}{u^2+1}$- $\displaystyle \int{\dfrac{2}{(u^2+1)\left( 7-\dfrac{10u}{u^2+1} \right) }}du$
- With simplification, you will get:
$\displaystyle 2\int{\dfrac{1}{7u^2-10u+7}}$- By completing the square, you will get:
$\displaystyle 2\int \dfrac{1}{\left( \sqrt{7}u-\dfrac{5}{\sqrt{7}} \right)^2+\dfrac{24}{7}}du$- Another substitution:
$\displaystyle s=\sqrt{7}u-\dfrac{5}{\sqrt{7}}$
$\displaystyle ds=\sqrt{7}du$- You will get:
$\displaystyle \dfrac{2}{\sqrt{7}} \int \dfrac{1}{s^2+\dfrac{24}{7}}$- The answer of this integral is:
$\displaystyle \dfrac{\arctan\left({\frac{1}{2}\sqrt{\frac{7}{6}} s}\right)}{\sqrt{6}}+C$
Where C is the integration constant.- You then go all the way back to substitute back the elements you used. For instance, you will plug into the resulting integral $\displaystyle s=\sqrt{7}u-\frac{5}{\sqrt{7}}$ so that you get a function of u. You will then substitute u for $\displaystyle u=\tan{\left(\dfrac{x}{2}\right)}$
The answer is:
$\displaystyle -\dfrac{\arctan{\left(\dfrac{5-7\tan{\frac{x}{2}}}{2\sqrt{6}}\right)}}{\sqrt{6}}+ C$
You magnificent to help this much I can offer 100 thanks but it only let me 1. I much pleased. I will not realise these step myself. But I hope one day I can help many people as you will help me. This is what I am much pleased for. You are gracious kindness much.
Much apology but how is sin(x) found to be as to be 2u/u^2+1?
Thank to you balloons, this is much inteligent to know this!
I have one more ask, it is how you are doing this integral to get arctan form
As I know d/dx arctan x = 1/x^2+1, but what here we have is not one it is 1/s^2 + 24/7
now i ask how this 24/7 will change the way you must to work on this step?
hope understand is clear what i ask
thankyou (i will work "24/7" and wont find this solution without the helps :P)
$\displaystyle \displaystyle{\frac{1}{s^2 + \frac{24}{7}}\ =\ \frac{1}{\frac{24}{7}(\frac{7}{24}s^2 + 1)}}$
$\displaystyle \displaystyle{\ =\ \frac{7}{24} \frac{1}{(\sqrt{\frac{7}{24}}s)^2 + 1}}$
$\displaystyle \displaystyle{\ =\ \frac{7}{24} \frac{1}{(\frac{1}{2}\sqrt{\frac{7}{6}}s)^2 + 1}}}$
no sorry that does not to help me, but much thank, i know you want help me,
i dont know how it relate to arctangent direvative.
as far i know only d/dx arctanx=1/x^2 + 1
i dont know it with other number but x^2 and 1, i dont know it when you have the co - eficient of x^2 or with another number than one
i know i would normal use my chainrules, but this not posible here
can you tell me is this also
d/dx arctan fx = 1/(fx)^2 + 1
can you only use this with a ONE and not another number
is it only posible, if the number term is made to 1
i ask reason is this, up to know, i have not see arctangent direvative. this is day 1 i try this piece... and is advance question.
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
Hope this helps to see how the chain rule is indeed the key. Others may post a more conventional working with Leibniz notation, but this way works for me.
NB of course I've only dealt with the 's' part of the journey.
_________________________________________
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Balloon Calculus; standard integrals, derivatives and methods
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I know the meaning from the balon, but it will not answer what i ask..
as i say i do not know the direvative arctan FX i can oonly if is just artan X. so i especialy dont kknow the rules for this. i especaily dont know to work it backward..
how u can integrate the 1/(x^2 + 24/7) i dont know this.
can some one please say, i will be very appreciate, what is definition direative of arctan fx
also you have put as constant rt7/2rt6
but should i be (2/rt7)*(7/24) equals as rt7/12
AS tom@ballooncalculus has already shown you:
$\displaystyle \displaystyle \int \frac{1}{x^2 + \frac{24}{7}}~dx$
$\displaystyle \displaystyle = \frac{7}{24} \int \frac{1}{\left ( \frac{7}{24} \right ) x^2 + 1}~dx$
Now let $\displaystyle y = \sqrt{\frac{7}{24}} x \implies x = \sqrt{\frac{24}{7}} y \implies dx = \sqrt{\frac{24}{7}}~dy$, so the integral becomes:
$\displaystyle \displaystyle = \frac{7}{24} \int \frac{1}{y^2 + 1} \cdot \sqrt{\frac{24}{7}}~dy$
$\displaystyle \displaystyle = \frac{7}{24} \cdot \sqrt{\frac{24}{7}} \int \frac{1}{y^2 + 1}~dy$
-Dan
that is not what i have ask
the change of x to y you post dan is not helpful, in fact, is not the method used earlier which i spend time understanding, you actually made thing much harder and unecesary, and out subsitution is x so term of y will not help
what i ask before (I CAN work it out if i know this... but no one answer this) is WHAT is direvative what arctan fx?
not arctan x but arctan fx how is chain rule use here when result is 1/x^2 + 1 because how it will affect on to the 1 in denom. i ask for general rule because i cannot find this from searching internet and is not in text book..
ALSO
when u are reaching the final answer
why is it u are putting
-arctan(5-7tan) instead of arctan(7tan-5)? why is the need to change this aspect becaue it is extrawork for the same event?
anyway the final step is to evaluate between limits 0 and 2pi?
please help, i make the definite integral ZERO because the tan(pi) and tan(0) are both zero the two terms cancel. how can i make it beetter
Well, I was deliberately ignoring the given constant of 2/rt7 and focussing on 1/(s^2 +24/7), but unfortunately I did mis-write that starting point (in the bottom left of the picture) as 1/(s^2 + 1), and this might well have confused you (or anybody) trying to make sense of the integration from the picture (now corrected). Try again working anti-clockwise from bottom left, noting that we're replacing
$\displaystyle \displaystyle{\frac{1}{s^2 + \frac{24}{7}}}$
right away with
$\displaystyle \displaystyle{\sqrt{\frac{7}{24}}\ \frac{1}{(\sqrt{\frac{7}{24}}\ s)^2 + 1}\ \sqrt{\frac{7}{24}}}$
which is the same as the re-arrangement given previously except for splitting 7/24 into root (7/24) times root (7/24).
Then note that we can replace the inner function [root(7/24) s] with tan theta, which results in an inner derivative of sec^2 theta, from which we can deduce that the top (integral) function F is arctan.
Now, you very sensibly want to be able to make sense of all this by looking at differentiating arctan of a function, using the chain rule. I was of course hoping that you would see the advantage of the diagram in doing just this - i.e. following it clockwise from top left instead of anti-clockwise from bottom left. But maybe my typo threw you off. Anyway, you will find plenty of examples of differentiating arctan of an inner function, but I hope you will look specifically at my similar examples at here and here.
thankyou much i understand how to get to the result now
but please some body emergency help me with this definite intigral limits
0 to 2pi
tom baloon you have confuse me a litle bit with the tan theta parts but what you have wirte in post #8 i can follow more well... i can use chainrule from this
but i need someone tell me is this now corrct
d/dx arctan fx will equal as (f'x)1/((fx)^2 + 1)
because what i did to to compare by this result which fernando post is divide the coficient of the integral by direvative of INSIDe this bracket (fx=(1/2)sqrt(7/6)x)
and i DID come to result which i will understand if and only if this general result i have assumed is actualy true?
tom i can folow your picture i undertaand what you have to doing in this but i dont know myself the rules formulas you have used and i want to apply it myself without just copy up the answer in to my study book... i want next time is not to have ask