I am tying to integrate $a_n= \frac{1}{L} e^x\sin\left(\frac{n\pi x}{h}\right)dx$ and get the solution in sinh form.  I have gotten the long answer, but cannot figure out how to turn it into sinh. Can someone help me with the steps?
120504
a) Compute the full Fourier series representation of $f(x) = e^{ax}, −π ≤ x &lt; π.$      b) By using the result of a) or otherwise determine the full Fourier series expansion for the function $g(x)=\sinh(x), −π ≤ x &lt; π.$   For part a) this is what I did and I'm pretty sure it's all correct. (Please let me know if it isn't!)  $$\begin{align} a_0\ &amp;=\frac{1}{π}\int_{-π}^{π} f(x)\cos(0x)\;dx\\ &amp;=\frac{1}{π}\int_{-π}^{π} e^{ax}\;dx=\frac{1}{π}\left[\frac{e^{ax}}{a}\right]^π_{-π}=\frac{e^{aπ}-e^{-aπ}}{aπ}\\ a_m\ &amp;=\frac{1}{π}\int_{-π}^{π} e^{ax}\cos(mx)\;dx\\ &amp;=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\sin(mx)}{m} dx\\ &amp;=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\cos(mx)}{m^2} dx\\ &amp;=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}+\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}a_m\\ a_m\left(\frac{m^2+a^2}{m^2}\right)&amp;=\frac{ae^{aπ}(-1)^m-e^{-aπ}(-1)^m}{πm^2}\\ a_m&amp;=\frac{(-1)^m\left(ae^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)} \end{align}$$  Then $$\begin{align} b_m&amp;=\frac{1}{π}\int_{-π}^{π} e^{ax}sin(mx)\;dx\\ &amp;=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\cos(mx)}{m}\;dx\\ &amp;=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\sin(mx)}{m^2}\;dx\\ &amp;=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}+\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}b_m\\ b_m\left(\frac{m^2+a^2}{m^2}\right)&amp;=\frac{(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{πm}\\ b_m&amp;=\frac{m(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)} \end{align}$$  so $$\begin{align} f(x)&amp;=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\ e^{ax}&amp;=\frac{e^{aπ}-e^{-aπ}}{π}\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right]\\ e^{ax}&amp;=\frac{2}{π}\sinh(aπ)\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right] \end{align}$$  but for part b) I wasn't sure of how to get the answer from part a) so instead I did this. I am also unsure if this is right. How do you do part b) using part a) ?  $$\begin{align} g(x)&amp;=\sinh(x)\\ a_0\ &amp;=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(0x)\;dx\\ &amp;=\frac{1}{π}\left[\cosh(x)\right]^π_{-π}=0\\ a_m\ &amp;=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(mx)\;dx\\ &amp;=\frac{1}{π}\left[\frac{\sinh(x)\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π}\frac{\cosh(x)\sin(mx)}{m}\;dx\\ &amp;=\frac{1}{π}\left[\frac{sinh(x)\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{\sinh(x)\cos(mx)}{m^2}\\ a_m\left(\frac{m^2+1}{m^2}\right)&amp;=\frac{1}{π}\left[\frac{\sinh(x)sin(mx)}{m}+\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}=0 \end{align}$$  Then $$\begin{align} b_m&amp;=\frac{1}{π}\int_{-π}^{π}\sinh(x)\sin(mx)\;dx\\ b_m\left(\frac{m^2+a^2}{m^2}\right)&amp;=\left[\frac{-\sinh(x)(-1)^{m}}{πm}\right]^π_{-π}\\ b_m&amp;=\frac{-2m\sinh(π)(-1)^m}{π(m^2+1)} \end{align}$$  so $$\begin{align} g(x)&amp;=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\ \sinh(x)&amp;=\frac{-2\sinh(π)}{π}\sum_{n=1}^{\infty}\frac{(-1)^nn\sin(nx)}{n^2+1} \end{align}$$
72909
How do we solve (i.e. get the closed form of) $\int \sqrt{1 + t^2} dt$ ?   shows the closed form of it but not the steps in solving it.  I think I need some algebraic trick..   I thought of some hyperbolic functions and their derivatives and equalities:  If I make $ t = \sinh x$, then I can reduce the integral to  $$ \int \cosh ^2 x dx $$ though I'm not sure if this is valid. But from here, I cannot go on evaluating the integral...
57798
$\int e^{2\theta}\ \sin 3\theta\ d\theta$  After Integrating by parts a second time, It seems that the problem will repeat for ever. Am I doing something wrong. I would love for someone to show me using the method I am using in a clean and clear fashion. Thanks.
81400
$$\sum_{n=1}^\infty \sin^{[n]}(1)$$ Where by $\sin^{[n]}(1)$ we mean $ \sin\left(\sin\left(\dots\sin(1)\right)\right)$ composed $n$ times.   Have tried the divergence test, which fails.  Have tried Ratio test, also fails, as the limit is 1.  Integral test, or root test do not seem promising. Help is appreciated
103744
So, I was recently working on the Sine Fresnal integral and was curious whether we could generalise for any Real Number, i.e.   $$I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx$$  I have formed a solution that I'm uncomfortable with and was hoping for qualified eyes to have a look over.   So, the approach I took was to employ Complex Numbers (I forget the name(s) of the theorem that allows this).   But   $$\sin\left(x^n\right) = \Im\left[-e^{-ix^n}\right]$$  And so, n  $$ I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx = \Im\left[\int_{0}^{\infty} -e^{-ix^n}\:dx \right]= -\Im\left[\int_{0}^{\infty} e^{-\left(i^{\frac{1}{n}}x\right)^{n}}\:dx \right]$$  Applying a change of variable $u = i^{\frac{1}{n}}x$ we arrive at:  \begin{align} I &amp;= -\Im\left[i^{-\frac{1}{n}}\int_{0}^{\infty} e^{-u^{n}}\:du \right] \\   &amp;= -\Im\left[i^{-\frac{1}{n}}\frac{\Gamma\left(\frac{1}{n}\right)}{n} \right]\\   &amp;= \sin\left(\frac{\pi}{2n}\right)\frac{\Gamma\left(\frac{1}{n}\right)}{n} \end{align}  My area of concern is in the substitution. As $i^{-\frac{1}{n}} \in \mathbb{C}$, I believe the limits of the integral should have been from $0$ to $i^{-\frac{1}{n}}\infty$. Is that correct or not?  I'm also struggling with bounds on $n$ for convergence. Is this expression valid for all $n\in\mathbb{R}$  Any guidance would be greatly appreciated
87616
In evaluating integral $\int\sin^{3}x dx $ I am pretty sure we need to use substitution $e^x=t$, but can't go  next step.
60447
I want to solve the integral $\int 1/(a^2+x^2)^n dx$. I guess this can be solved using the reduction formula as in the case of $\int x^2/(a^2+x^2)^n dx$. However, I am not able to do it.
10157
I have stumbled onto an interesting integral$$\int_0^\infty \sin(x)\sin\left(\frac{1}{x}\right)dx$$ which I noticed graphically that it appears to be $1$, but I have no idea on how to evaluate it. Maybe it could be done with the use of Bessel Functions? Any help is appreciated.
75214
How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?  It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
93963
I don't know how to integrate $\int x^2\sin^2(x)\,\mathrm dx$. I know that I should do it by parts, where I have  $$ u=x^2\quad  v'=\sin^2x \\  u'=2x \quad v={-\sin x\cos x+x\over 2}$$  and now I have  $$ \int x^2\sin^2(x)\,\mathrm dx = {-\sin x\cos x+x\over 2} x^2 - \int 2x{-\sin x\cos x+x\over 2}\,\mathrm dx\\ ={-\sin x\cos x+x\over 2} x^2 - \int x(-\sin x\cos x+x)\,\mathrm dx\\  $$  so I have to calculate $$ \int x(-\sin x\cos x+x)\,\mathrm dx=-\int x\sin x\cos x\,\mathrm dx+\int x^2\,\mathrm dx$$  I know that $\int x^2\,\mathrm dx = {1 \over 3}x^3+C$ but I don't know what should I do with $$\int x\sin x\cos x \,\mathrm dx$$ Should i use parts again or what? Please help.
52255
Solve the indefinite Integration $\int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx$   $$ \int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx=\int \frac{1}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx\\=\frac{1}{\sin\alpha}\int \frac{\sin(x+\alpha-x)}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx=\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos x-\cos(x+\alpha)\sin{x}}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx\\ =\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos x}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx-\frac{1}{\sin\alpha}\int\frac{\cos(x+\alpha)\sin{x}}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx $$  Is it possible to proceed further and complete the integration or what is the right substitution to find the solution ?  Note: I'm looking for a simple way to solve this, unlike here .
69336
Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:  Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.   Taking the derivatives: $f^{(n)}(x) = \dfrac{1}{2i} \cdot ((1+i)^n e^{(1+i)x} - (1-i)^n e^{(1-i)x})$  Now, I use that: $$ (1+i)^n = {\sqrt{2}}^n \cdot \left(\cos\dfrac{n \pi}{4} + i \sin\dfrac{n \pi}{4}\right) \\(1 - i)^n = \sqrt{2}^n \cdot \left( \cos \dfrac{-n \pi}{4} + i \sin \dfrac{-n \pi}{4} \right)$$  Plugging that mess, I get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\left(\cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4}\right) e^{ix} - \left(\cos \dfrac{-n \pi}{4} + i \sin \dfrac{- n \pi}{4}\right) e^{-ix} \right)$$  But, $e^{ix} = \cos x + i \sin x$, and using Moivre's theorem, that makes: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\cos \left( x + \frac{n \pi}{4}\right) + i \sin \left( x + \frac{n \pi}{4}\right) - \left(\cos \left( - x  - \frac{n \pi}{4}\right) + i \sin \left( -x -\frac{ n \pi}{4}\right)\right)\right)$$  and since $\cos$ is an even function, and $\sin$ is odd, we get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot 2i \sin \left(x + \dfrac{n \pi}{4}\right)$$  Simplifying, the answer would be $f^{(n)}(x) = e^x \cdot \sqrt{2}^n \cdot \sin\left(x + \dfrac{n \pi}{4}\right)$. I'm almost positive that this is it, but I just want to be sure. Thank you in advance!
28624
Calculate the integral for $\left|a\right|&lt;1$  $$\int_{-\pi}^{\pi} \dfrac{\cos(n\theta)}{1-2a\cos(\theta)+a^2}d\theta$$  I'm supposed to evaluate this using method of residues, but the parameter a is confusing me. I'm not even sure what contour I should use. Any help would be very appreciated.
44339
I'm having trouble calculating the integral  $$\int_{- \infty}^\infty \frac{\sin x}{x+i}\,dx $$  using residue calculus. I've previously encountered expressions of the form   $$\int_{- \infty}^\infty f(x) \sin x \,dx $$  where you would consider $f(z)e^{iz}$ on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with $e^{iz}$ in my complex function because $$\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}\cdots$$  How to remedy this? I'm not sure if substituting $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work
84518
Integrate the following  :   $$\int \frac{\sin x}{\sin4x}dx$$  My approach :   $$=\int \frac{\sin x}{2\sin2x \cos2x}dx$$  $$= \int \frac{\sin x}{4\sin x \cos x \cos2x}dx$$ $$= \int \frac{1}{4\cos x \cos2x}dx $$ [ didn't get the hint here ] then I did the following way.  $$= \int \frac{\sin x}{4\sin x \cos x (\cos2x) }dx$$  Now putting $\cos x =t$ $\Rightarrow -\sin x \, dx = dt $   $$= \int \frac{-dt}{4\sqrt{1-t^2} t(2t^2-1)}$$  [Using : $ \cos2x = 2\cos^2x-1]$
85114
I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.  Any suggestions are welcome. Thanks.
68284
closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$  for $a,b&gt;0$  n=1 we get  $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$  n=2 we get  $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{a^3b^3}$$  but what the integral for n ????  I hope to be there two different solution one by contour integration and the other by real analysis and thanks for all