Use integration by parts to calculate 𝐼 =   f  e^(ax) cos 𝑏𝑥 𝑑𝑥 for 𝑎 ≠ 0 and 𝑏 ≠ 0.  This is my working so far. The part where i have to integrate is confusing so with some help maybe i can do it .
104960
I am curious about the proof of these two integrations. It would be great if you could point me in the right direction : $ \int{e^{ax}\cos({bx})dx}= \dfrac{e^{ax}}{a^{2}+b^{2}}[a\cos(bx)+b\sin(bx)] + C $ &amp; $ \int{e^{ax}\sin({bx})dx}= \dfrac{e^{ax}}{a^{2}+b^{2}}[a\sin(bx)-b\cos(bx)] + C $
36902
How to find integral of: $$\int xe^x\sin(x)$$ I know it should be repeated integration by parts, but i do not know how many times i should do this and when to stop.
69895
I am supposed to integrate $f(z)=$$\frac{5}{z}$ from -3 t0 3 but I am having trouble understanding how to do this. I've done the integration the "hard" way by using parametrizations but now I need to use the fact that $f(z)$ is analytic everywhere except at $z=0$ and a different method to perform the integration but I am stumped.
55214
The Problem is to integrate the following double integral $\displaystyle\iint_\mathbf{D}\exp\left(\frac{x-y}{x+y}\right) dx\, dy$ using the technique of transformation of variables ($u$ and $v$).  There is a given $D$ with $D:= \{(x,y):\ x\geq 0,\ y \geq 0,\ x+y \leq 1\}$  My Approach: wasn't quite successful at all. I am quite familar with calculus, but i am stuck at multiple integrals. I thought about the following transformation:  $u = x-y$ and $v = x+y$  But i don't know the next step.  Another suggestion: i thought about the integration of the e-Function. I have in mind that $\int e^x dx = e^x$. So maybe: $$\iint_\mathbf{D} \exp\left(\frac{x-y}{x+y}\right) dx\ dy = \exp\left(\frac{x-y}{x+y}\right) ?$$
63713
This  is my problem, true or false?  prove that:   $$\int_{0}^{\infty}\frac{\displaystyle{\sum_{k=1}^{\infty}}k\sin(kx)\,e^{-tk^2}}{\displaystyle{\sum_{k=1}^{\infty}}\cos(kx)\,e^{-tk^2}}dt=\frac{\pi^2({\pi-x})}{8}$$ and $0&lt;x&lt;\dfrac{\pi}{2}$  Thank you.
81210
As the title states, I'm having quite a lot of trouble with the integral:  $$\int_{-\ln 2}^{\ln 2}  \frac{(\sin x+x)^{1/3}}{e^x}dx$$  The problem is that my standard (admittedly sparse) repertoire of tricks seems to have no effect at all. I can't think of any clever substitution that would work, and it looks like a nasty integral all-round.   I was hoping to apply some symmetry arguments (especially considering the limits of integration), but the exponential function is neither even nor odd, so I don't see how this would work.   I'm afraid I have absolutely no idea how to approach this.
63223
Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$  How can I evaluate it? Thanks!
81166
How to evaluate the integral $$\int e^{x^3}dx \quad ?$$  I've tried to set $t=x^3$, but it seems to be a blind alley; I don't know what to do with $\int\frac{e^t}{3\sqrt[3]{t^2}}dt$.
123576
Given $$\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$$ evaluate:   $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx. $$   I can find that  $$\left(ax+\frac{b}{x}\right)^2 = a^2x^2+2ab+\frac{b^2}{x^2}$$  therefore: $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx = e^{2ab}\int_0^{\infty}e^{-\left(ax+\frac{b}{x}\right)^2}dx$$  but I can't find any clue then.
34118
$$\int_{0}^{\frac{\pi }{2}}x\cot(x)dx$$   I tried integration by parts and got $\dfrac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} \csc^{2}x dx$ which doesn't help at all. I don't really know what to do. Any help will be greatly appreciated.
30762
Question:     Find out $\displaystyle{\int e^{\cos x}~dx}$.   My Attempt:  Let $\cos x = y$. Hence $-\sin x\ dx = dy$ or $$dx = \displaystyle{\frac{-dy}{\sin x}=\frac{-dy}{\sqrt{1-\cos^2x}}=\frac{-dy}{\sqrt{1-y^2}}}$$ So  $$\begin{align}\int e^{\cos x}~dx &amp;= \int e^y\left(\frac{-dy}{\sqrt{1-y^2}}\right)\\ &amp;=-\int\frac{e^y}{\sqrt{1-y^2}}~dy \end{align}$$  This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to solve. I can't find out any substitution that I can make in this integral to make it simpler. Please help me solve it. Is the problem with my first substitution $y=\cos x$ or is there any other way to solve the integral $\displaystyle{\int\frac{e^y}{\sqrt{1-y^2}}~dy}$?
53688
I'm trying to solve this integral: $$\int_{-\infty}^\infty e^{-x^2-x}dx$$ WolframAlpha shows this to be approximately $2.27588$. I tried to solve this by integration by parts, but I just couldn't get there. I'd be glad if someone could show me how to do it.   I've included my attempt at the problem below:  Note $\int_{-\infty}^\infty e^{-x^2-x} = \int_{-\infty}^\infty e^{-x^2}e^{-x}dx$. Then we can integrate by parts. Let $u(x) = e^{-x^2}$. Then $u'(x) = -2xe^{-x}$. Let $v'(x) = e^{-x}$. Then $v(x) = -e^{-x}$.  Then $u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)dx$, i.e. $$\int_{-\infty}^\infty e^{-x^2}e^{-x}dx = -e^{-x^2}e^{-x} - \int e^{-x}2xe^{-x}dx = -e^{-x^2-x} - 2\int e^{-2x}xdx$$  Then we integrate $\int e^{-2x}xdx$ by parts. We pick $u(x) = x$, $u'(x) = dx$, $v'(x) = e^{-2x}$, and $v(x) = \int e^{-2x} dx = \frac {-1}{2} e^{-2x}$ by u-substitution. Skipping some steps, it follows that  $$\int e^{-2x}xdx = -\frac{1}{4}(e^{-2x})(2x+1)$$ Then  $$ -e^{-x^2-x} - 2\int e^{-2x}xdx = -e^{-x^2-x} + \frac{1}{2}(e^{-2x})(2x+1)$$ Which, when evaluated numerically, doesn't yield the desired result. So there's a problem here. Maybe someone else knows how to do this.
102762
I want to integrate  \begin{align} \int \frac{ev+f}{av^2 + bv +c} dv \end{align} can you give me some hints or detail procedure for this integral?    From mathematca, i have  \begin{align} \frac{-\frac{2 (b e-2 a f) \text{ArcTan}\left[\frac{b+2 a v}{\sqrt{-b^2+4 a c}}\right]}{\sqrt{-b^2+4 a c}}+e \text{Log}[c+v (b+a v)]}{2 a} \end{align}  which seems uncomfortable for me..
9190
I'm working on the Fourier transform, but I don't know how to evaluate the integral:  $$I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $$
40732
$$\frac{dy}{dx} = x-y$$  How do I separate so I can integrate both sides? Thanks for getting me started.  I know the solution is $$ y = x-1+2e^{-x}$$
51644
I know that $\int{\frac{1}{x}}dx$ is simply $\ln{(x)}+c$ (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach $e^{x{^2}}$. Perhaps a substitution?
91174
I was wondering how to determine $$\int x^2 e^{-x^2}\mathrm dx?$$ I tried to move the exponential into the integrator, and apply integration by parts, but that will make the integrand more complicated. Thank you!