2 votes
Accepted

Need help with the Inverse Laplace transform

Convolution theorem: $$\mathcal L(f(t)*f(t))=(\bar f(s))^2$$ Here $$\bar f(s)=4/(s^2+4)$$ So \begin{aligned} f(t)=\mathcal{L}^{-1}(\bar{f}(s)) & =\mathcal{L}^{-1}\left(\frac{4}{s^2+4}\right) \\ &...
SHBooKP's user avatar
  • 194
2 votes

definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$

Note that \begin{align} I=&\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx\\ =&\int_{0}^{\frac{\pi}{4}} \frac{(\sin x\cos x)^2(\sin x+\cos x)}{(\sin x+\cos x)^2(\sin^2x-\sin ...
Quanto's user avatar
  • 97.5k
1 vote

Help request for calculating an integral

Alternatively, let $x=\frac t{\sqrt{1+t^2}}$ \begin{align} \int \frac{2\sqrt{1- x^2}}{2 x\sqrt{1- x^2}+ 5}dx =& \int \frac{2}{(1+t^2)(5t^2+2t +5)}dt\\ =&\ \int \frac{t+\frac25}{t^2+\frac25 t +...
Quanto's user avatar
  • 97.5k
1 vote

Help request for calculating an integral

Letting $x=\sin u $ we get: $$ \int \frac{2\cos^2 u}{2\sin u \cos u+5} du $$ Note that $2\cos^2 u =\cos(2u)+1$ and $2\sin u \cos u =\sin(2u)$ $$ \int {\cos(2u)+1\over\sin(2u)+5} du = \frac{1}{2}\int \...
Masd's user avatar
  • 567
1 vote
Accepted

Help request for calculating an integral

Your initial simplification is wrong. Implementing the trig substitution right away, while assuming $\cos y>0$ for brevity, yields $$\int \frac{2\sqrt{1- x^2}}{2 x\sqrt{1- x^2}+ 5} \, dx \stackrel{...
user170231's user avatar
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1 vote

Integral of $e^{-x^2}$

First please check out: Is there really no way to integrate $e^{-x^2}$? There is no form of the integral in terms of elementary function. But you can use infinite series: $$ \int_a^b e^{-x^2}dx = \...
Masd's user avatar
  • 567
1 vote
Accepted

Why doesn't $\iiint\limits_\Omega (y^2+z^2)dv$ equal $\int_0^5 2x\pi\cdot 2x\,dx$, rotating the curve $y^2=2x$ around the x-axis and the plane $x=5$?

The right answer is indeed $\frac{250}{3}\pi$. I'm going to demonstrate why is that and then discuss why your approach isn't what you usually want to do. So, the shape we're given is actually a ...
Egor Larionov's user avatar
1 vote
Accepted

Prove that $f$ doesn't attain its local maximum

As @RyszardSzwarc pointed out in his comment, if there was a point $(x_0,y_0)$ in which the function has a local maximum, then the two functions of one variable $$ g(x) = f(x,y_0), \qquad h(y) = f(x_0,...
Vincenzo Tibullo's user avatar
1 vote

What is $\int (y'(x))^2 dx$?

It is true that you cannot do this integral without knowing what $y$ is. But we can simplify it: $$ \int (y'(x))^2 dx = \int y'(x)y'(x) dx=y(x)y'(x)-\int y''(x)y(x)dx $$ There isn't much we can do ...
Masd's user avatar
  • 567
1 vote

Find Maximum and Minimum distance from origin to $f(x,y)$ using the Lagrange method.

You've done fine to conclude that $x^2=y^2.$ That means $x=y$ or that $x=-y.$ Plug each of those into your curve's equation. You should find four distinct points of two distinct distances from the ...
Cameron Buie's user avatar
1 vote
Accepted

antiderivative of $\cos^nx \cos(nx)$

Another straightforward way is to use $cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$. We get $\cos^n x \cos(nx)=\frac{1}{2^{n+1}}(e^{ix}+e^{-ix})^n(e^{inx}+e^{-inx})=\frac{1}{2^{n+1}}((e^{2ix}+1)^n+(1+e^{-2ix})^...
Vit's user avatar
  • 321
1 vote

Difficult Vectors Problem (Calculus & Vectors)

$ \def\R#1{{\mathbb R}^{#1}} \def\q{\quad} \def\qq{\qquad} \def\qiq{\q\implies\q} \def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\A{{\bar\a}} \def\B{{\bar\b}} \def\G{{bar\g}} \def\l{\lambda} ...
greg's user avatar
  • 35.9k
1 vote

Find $\int_{0}^{\pi/2} \frac{\mathrm{d}\theta}{(\sin\theta+1)^2}$.

Use $u=x+\sqrt{1+x^2} $(for the original integral, this substitution is also not a bad choice), then $$x=\frac{1}{2}(u-1/u),\quad {\rm d}x=\frac12(1+1/u^2){\rm d}u.$$ So $$\int_{0}^\infty \frac{{\rm d}...
Riemann's user avatar
  • 7,245

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