# Thread: HOW DO I DO THIS QUESTION: Integration

1. ## HOW DO I DO THIS QUESTION: Integration

So recently while reading BNHA Manga, I realise this question:

Integration of (0 to ln(1+sqrt2))((e^x-e^-x)/2)^3 . ((e^x+e^-x)/2)^11

No clue pls help! Amswer 107/24

2. ## Re: HOW DO I DO THIS QUESTION: Integration

Originally Posted by Denki
So recently while reading BNHA Manga, I realise this question:

Integration of (0 to ln(1+sqrt2))((e^x-e^-x)/2)^3 . ((e^x+e^-x)/2)^11

No clue pls help! Amswer 107/24
$\noindent I'll give you an outline of how to compute the \emph{in}definite integral I = \int \left(\frac{e^{x} - e^{-x}}{2}\right)^{3} \left(\frac{e^{x} + e^{-x}}{2}\right)^{11} \, dx. If you can do this, you should be able to do your question.$

$\noindent Write \color{blue}\mathcal{S} = \frac{e^{x} - e^{-x}}{2}\color{black} and \color{blue}\mathcal{C} = \frac{e^{x} + e^{-x}}{2}\color{black}. Then I = \int \mathcal{S}^{3} \mathcal{C}^{11}\, dx. Write I = \int \mathcal{S}^{2} \mathcal{C}^{11}\times \mathcal{S}\, dx. Observe the following (easy-to-prove) facts:.$

$\noindent 1) \mathcal{S}^{2} = \mathcal{C}^{2} - 1$
$\noindent 2) \mathcal{C}' = \mathcal{S}.$

$\noindent Hence I = \int \left(\mathcal{C}^{2} - 1\right) \mathcal{C}^{11}\, \mathcal{C}' \, dx. Hence substituting u = \mathcal{C} (so du = \mathcal{C}'\, dx), we have I = \int \left(u^{2} - 1\right) u^{11}\, du = \cdots. (Final answer for the indefinite integral should be re-expressed in terms of x, using the definition of \mathcal{C}.)$

3. ## Re: HOW DO I DO THIS QUESTION: Integration

$\noindent By the way, the functions \mathcal{C} and \mathcal{S} above are known as the \textit{hyperbolic functions} \cosh and \sinh. They have a number of interesting properties (for example, each is the derivative of the other) and are related to the trigonometric functions \cos and \sin. You can find out more about them at \color{blue}\texttt{https://en.wikipedia.org/wiki/Hyperbolic\_function}\color{black}.$

4. ## Re: HOW DO I DO THIS QUESTION: Integration

LOL, Ths is from the januray issue. i remeber i was tryna figure it out too LOL
i asked my teacher and he did it fir me

Carrying on wat integrand said,

sinh^-1x is ln(1+sqrt(x^2+1))

if u put 1 as x then its the same as ln|1+sqrt(2)|

and sinh(sin^-1(1))=1

simplifies it a lot

if u need the wrking out i can do it fir u

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