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0

The second approach fails because $\sin x+\cos x$ is not a bijective function on $[0,\pi/2]$. If you like a third approach: $$I=\int_{0}^{\pi/2}\left(\sin x-\cos x\right)\log \sin x\,dx = \left.\frac{d}{d\alpha}\int_{0}^{\pi/2}\left(\sin^{\alpha+1}x-\cos x\sin^{\alpha} x\right)\,dx\,\right|_{\alpha=0}$$ leads to: $$\begin{eqnarray*} I &=& ...


2

The equation of your curve rewrites as follows:(just complete the square) $$x^2+\left(y-\frac{a}{2}\right)^2=\frac{a^2}{4}$$ Here's a circle centered at $(0,\frac{a}{2})$ of radius $\frac{a}{2}$ A parametrisation of that circle is $$\begin{cases} x=\frac{a}{2}\cos\theta\\y=\frac{a}{2}\left(1+\sin\theta\right)\end{cases}$$ Can you take it from here?


2

Hint: $$x^2 + y^2 - ay = x^2 + (y-\frac a2)^2 - \frac{a^2}{4}$$ Meaning that $\Gamma$ is the circlle defined by the equation $$x^2 + (y-\frac a2)^2 = \left(\frac{a}{2}\right)^2$$


1

If $a\ne0,$ $$x^2+\left(y-\dfrac a2\right)^2=\left(\dfrac a2\right)^2\implies\left(\dfrac x{\dfrac a2}\right)^2+\left(\dfrac{y-\dfrac a2}{\dfrac a2}\right)^2=1$$ Use $\cos^2\theta+\sin^2\theta=1$


0

The using of method "substitution" needs $t$ changes from $\alpha=\phi(a)$ to $\beta=\phi(b)$ monotonically while $x$ changs from $a$ to $b$ monotonically.


2

You cannot use that substitution. When you substitute, you are actually composing with a function $x=\varphi(t)$ in this way: $$\int_{\varphi(a)}^{\varphi(b)}f(x)dx=\int_a^bf(\varphi(t))\varphi'(t)dt$$ Can you write $\sin(x)+\cos(x)=t$ as $x=\varphi(t)$? The answer is no: the function $t=f(x)=\sin(x)+\cos(x)$ is not invertible in $[0, \pi/2]$ because it ...


2

No, there isn't. As long as you can apply the integral test, then it has to work: if $f : [0, \infty) \to [0, \infty)$ is decreasing, then the series $\sum f(n)$ converges iff the improper integral $\int_0^\infty f(t) dt$ converges. This isn't, say, like the root test or the ratio test that can be inconclusive, even when you can apply it (when $\lim ...


0

The answer is $\frac{58}{15}$. For questions like these, a readily available resource is WolframAlpha (which is what I used). The tricky part, I would guess, is finding the bounds of the integrals, which is either $\int_0^2 \int_{\frac{1}{2}y}^1$ or $\int_0^1 \int_0^{2x}$ depending on your chosen order of integration.


1

Define $$I_f(k) := \sum_{i=1}^{k} h(f_i, f_{i+1})$$ Then $$H(f)_k = f_{\max \{ i \mid I_f(i) \le k \tau_t \}} \qquad \text{for } k \le \lfloor I_f(n) / \tau_t \rfloor$$ Where $H(f)_k$ is the $k$-th output, i.e. $H(f)$ is a sequence of length $\le n$.


2

You should know that using the residue theorem would be easier, but if we are to restrict ourselves to only Cauchy's integral formula, then here's one way of attacking it: First, note that the integrand is a quotient of two entire functions. As such, the integrand is analytic everywhere except the points at which the denominator is zero. Since $z^2+4$ can ...


1

Since all the poles of the integrand are enclosed by the contour, we have \begin{eqnarray} \int_{|z|=3}\frac{e^{zt}}{z^2+4}dz=2\pi i\text{Res}_{z=0}\frac{1}{z^2}\frac{e^{\frac{t}{z}}}{\frac{1}{z^2}+4}=2\pi i\text{Res}_{z=0}\frac{e^{\frac{t}{z}}}{1+4z^2} \end{eqnarray} Note that $\displaystyle \frac{e^{\frac{t}{z}}}{1+4z^2}=\left(\sum_{n=0}^\infty ...


1

Hint. The denominator $\displaystyle z^2+4=(z-2i)(z+2i)$ of the function $$ z \longmapsto \frac{e^{zt}}{z^2+4} $$ gives two poles inside $|z|=3$ and by the Cauchy integral formula we have $$ \int_{|z|=3}\frac{e^{zt}}{z^2+4}dz=2i\pi\left({\rm{Res}}_{z=-2i}f(z)+{\rm{Res}}_{z=2i}f(z)\right). $$ You conclude using for example $$ {\rm{Res}}_{z=-2i}f(z)=\lim_{z\to ...


2

Write integral in the form the function in the form $\frac{e^{zt}}{(z-2i)(z+2i)}$. Then you should split the contour in two parts such that interior of each part contain only one point $2i$ or $-2i$. Then apply Cauchy integral formula for each contour.


1

The bounds of the integral are being passed to (12-y), so smaller y results in larger areas of integration. The first equation seems easier to "visualize" to me too, but I think the second one can be intuited as well. The first term is the work done for the elevator. The second term is the work done for the cable as it moves from position zero to position ...


0

Let $f$ be Riemann integrable on $[a, b]$. Then the function $$F(x) = \int_{a}^{x}f(t)\,dt$$ is known to be continuous and of bounded variation on $[a, b]$. Apart from these properties of $F$ it is possible to ensure differentiability of $F$ at points where $f$ is continuous. Another point to note is that since $F(x)$ is of bounded variation it possesses a ...


1

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx$$ We can write $$\displaystyle (x\sin x+\cos x) = \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \sin x+\frac{1}{\sqrt{1+x^2}}\cdot \cos x\right\} = \sqrt{1+x^2}\cdot \cos \left(x-\phi \right)$$ So Here $$\displaystyle \cos x\ \phi = \frac{1}{\sqrt{1+x^2}}$$ and ...


4

Consider a square such that the distance from the centre to any side is $r$. Then the area of the square is $4r^2$, and the perimeter of the square is $8r$, which is the derivative of $4r^2$. So your circle rule works for the square, if we use the right parameter to describe its size. Exploration: Let us play the same game with the equilateral triangle. ...


4

Denote your integral by $I$. Note that $f(z) = z^r(z+\lambda)^{-1}$ has branch points at $z=0$ and $z=\infty$ (from the $z^r$, given that $r\in (-1,0)$), and that two cases arise depending on the value of $\lambda$: (1) a simple pole at $z=-\lambda$ if $\lambda >0$ (2) no poles if $\lambda = 0$ But note that the real integral does not converge for case ...


0

Here's a more elementary way of seeing it from the derivative of $\sec{\theta}$: $\frac{\sec(\theta+\delta)-\sec\theta}{\delta}=\frac{\frac{1}{\cos(\theta+\delta)}-\frac{1}{\cos\theta}}{\delta}=\frac{cos\theta-\cos(\theta+\delta)}{\delta\cos\theta\cos(\theta+\delta)}$, So $\frac{d}{d\theta}(\sec{\theta})=\lim\limits_{\delta\rightarrow ...


3

The expression $\sec x\tan x$ can be written $$ \frac{1}{\cos x}\frac{\sin x}{\cos x}=\frac{\sin x}{\cos^2 x} =-\frac{-\sin x}{\cos^2 x}=-\frac{f'(x)}{f(x)^2} $$ where $f(x)=\cos x$. Consider, for a generic differentiable function $f$, $$ g(x)=\frac{1}{f(x)}. $$ By the chain rule $$ g'(x)=-\frac{f'(x)}{f(x)^2}. $$ In the special case of $f(x)=\cos x$, we see ...


6

Lets consider the indefinite integral, $$ \int \sec(x)\tan(x) dx = \int \frac{\sin(x)}{\cos^2(x)} dx$$ We can then perform a $u$ substitution with $u=\cos(x)$ and $du = -\sin(x) dx $ obtaining, $$ \int \sec(x)\tan(x) dx = -\int \frac{1}{u^2} du= \frac{1}{u} + C = \frac{1}{\cos(x)} +C = \sec(x) + C$$


6

Hint: Here's something for you to think about. By the Chain Rule $$\frac{d}{dx} \left(\cos (x)\right)^{-1} = (-1) (\cos (x))^{-2} \dot \, (-\sin (x)) = \frac{1}{\cos x} \dot\, \frac{\sin x}{\cos x}$$ Edit: In order to find $\int \sec x \tan x dx $. Write it as $$\frac{\sin x}{\cos^2 x}$$ and make the substitution $u = \cos x$.


0

I have answered it in the paper , I think that it's correct


0

i think the first integral must be $$\int_{2/3}^{1}(3x-2)dx$$


2

Let $\varepsilon>0$, then since $\int_{\Omega}|X|dP<\infty$, there exists an $n>0$ such that: $$nP(|X| > n) = \int_{\{\omega\in\Omega : |X(\omega)|>n\}}ndP \leq \int_{\{\omega\in\Omega : |X(\omega)|>n\}}|X(\omega)|dP < \varepsilon.$$ Therefore we have $$nP(|X| > n) < \varepsilon.$$


2

$$\int_\Omega |X|\, dP < \infty \implies \int_{\{|X| > n\}} |X|\, dP \to 0$$ by the dominated convergence theorem. Since $nP(|X|>n)$ is bounded above by the last integral, we have it.


2

Hint: To simply the notation define $\ln_k(x) \equiv \ln\ln\cdots\ln(x)$ where there are $k$ logs on the right hand side. With this notation your integral can be written $$\int^\infty\frac{dx}{x\ln_1(x)\ln_2 (x)\cdots \ln_{k-1}(x)(\ln_k(x))^s}$$ From the definition above it follows that $\ln_{k}(x) = \ln(\ln_{k-1}(x))$ so $$\frac{d\ln_{k}(x)}{dx} = ...


0

METHOD 1: First note that we can write $\vec E(\vec p)=-\nabla \Phi(\vec p)$. We also establish a coordinate system so that $\hat p$ is the polar axis. Thus, we can simplify the integral substantially by writing $$\begin{align} \vec E(\vec p)&=\frac{1}{4\pi \epsilon_0} \int_V \rho(\vec q) \frac{\vec p-\vec q}{(p^2+q^2-2\vec p\cdot \vec q)^{3/2}} ...


1

The easiest way is to apply the long division, but you can make substitution also: let $u=x^2+1\Rightarrow du=2x dx$ Therefore: $\int \frac{x^3+2x}{x^2+1}dx=\frac{1}{2}\int\frac{u+1}{u}du=\frac{u+ln(u)}{2}+C=\frac{x^2+1+ln(x^2+1)}{2}+C=\frac{x^2+ln(x^2+1)}{2}+C$ I don't understand if @mathlove is the first who gives an answer why others give the same ...


5

To integrate a rational function there are two standard methods: the partial fractions decomposition, and the Ostrogradski-Hermite method. The latter does not require finding the roots of the denominator. See example here.


5

The first step is generally to divide out the integrand so as to get a polynomial plus a rational function whose numerator has lower degree than its denominator. Here you get $$\frac{x^3+2x}{1+x^2}=x+\frac{x}{x^2+1}\;.$$ Integrating the $x$ term (and, more generally, the polynomial quotient) is easy, so we’ve reduced the problem to integrating something of ...


2

Generally, you just try the long division if degree of $a(x)$ is higher than degree of $b(x)$. And for this particular example, write $\dfrac{x^3+2x}{1+x^2} = x + \dfrac{x}{1+x^2}$. Can you take it from here?


1

Hint: $x^2+y^2\, dV = r^2\,r\,dr\,d\theta\,dz=r^3\,dr\,d\theta\,dz$ in polar coordinates


2

For odd values of n, use the fact that $~\dfrac n{(1+x)(1+x^n)}~=~\dfrac1{(1+x)^2}~+~\dfrac{P_{n-2}(x)}{1+x^n}~,~$ where $P_{n-2}(x)~=~\displaystyle\sum_{k=0}^{n-2}(n-k-1)(-x)^k,~$ in conjunction with the famous identity $~\displaystyle\int_0^\infty\frac{x^{a-1}}{1+x^n}~dx$ $=~\dfrac\pi n\cdot\csc\bigg(a~\dfrac\pi n\bigg),~$ which can be proven using the ...


0

There is no general rule in introducing an extra variable to apply the Feynman differentiation trick. In this case it looks reasonable to consider $\arctan(x)$ as $\arctan(ax)|_{a=1}$ because the derivative of the arctangent is a rather simple algebraic function. So, let: $$ I(a) = \int_{0}^{1}\frac{\arctan(ax)}{\sqrt{1-x^2}}\,dx. \tag{1}$$ We trivially have ...


1

You forgot a $(-)$ in an intermediate step but somehow you corrected it in the next step. Your final result is correct. It will not be simplified further (as far as I can see) $(ii)$ To find the coefficients you had $$\frac16 \int ^3 _ 1 12 e^{-in\frac{\pi}{3}x}dx$$ Notice that the integral you want is $n=-3$ (with a factor of $2$ in front of it)


3

$a)$ One thing you need to know by heart is this transformation: $x^2+y^2=r^2$, so the integrand becomes $\ln (1+r^2)$. We also have $dx\,dy=rdr\,dt$ (I will use $t$ instead of $\theta$ because it is easier for me) To compute the integral in the unit circle, you need to consider not $r=1$ but $r<1$ and $0<t<2\pi$ (they should have taught you this, ...


1

Either you know Riemann-Lebesgue's lemma, and it asserts such a limit is $0$, or you can prove it with in integration by parts: Set $u=\dfrac1{\cos t+4}$, $\mathrm d\mkern1mu v=\cos nt \mathrm d\mkern1mu t$. You obtain: $$\int_{0}^{2\pi} \frac{\cos nt} {\cos t+4}\mathrm d\mkern1mut=\frac{\sin t}{n(\cos y+4)}\Biggr\rvert_0^{2\pi}-\frac1n\int_{0}^{2\pi} ...


0

$1.$ You can find the height of the cylinder by using Pythagorean Theorem. Draw the radius from the center to the edge of the cylinder and notice that $\text{height}=2\sqrt{R^2-\rho^2}$. Now you can subtract this cylinder and the two spherical caps to find the volume. I will not derive the formula for spherical cap volume here. You can look here: ...


1

We have $\displaystyle\int_0^{\infty}\dfrac{dx}{(1+x)(1+x^n)}=\int_0^{\infty}\dfrac{dx}{1+x^n}-\int_0^{\infty}\dfrac{xdx}{1+x^n}$ $=\displaystyle\frac{1}{n}(B\Bigl(\frac{1}{n},1-\frac{1}{n},\Bigr)-B\Bigl(\frac{2}{n},1-\frac{2}{n},\Bigr))=\dfrac{\pi}{n\sin(\pi/n)}-\dfrac{\pi}{n\sin(2\pi/n)}$. Second try. We have ...


0

Hint The subset of $K$ whose volume you are computing can be described in cylindrical coordinates as $$K \cap \{(r, \theta, z)_c : r > \rho \}.$$ (It is unfortunate here that the symbol, $\rho$, which in this setting is often reserved for the [co]latitudinal spherical coordinate is used in even one other way here, let alone two.)


0

Note that we can write $\cos^n(x)=\cos(x)\cos^{n-1}(x).$ Use the formula of integration by parts \begin{equation*} \int udv=uv-\int vdu \end{equation*} with $u=\cos^{n-1}(x)$ and $dv=\cos(x)$ to get \begin{equation*} \int\cos^n(x)=\int \cos^{n-1}(x)\cos(x)dx=\sin(x)\cos^{n-1}(x)--(n-1)\int\cos^{n-2}(x)\sin(x)\sin(x)dx \\ = ...


1

Suppose we seek to evaluate $$\frac{1}{2} \int_0^{2\pi} \frac{1}{a+b\sin^2 x} dx.$$ Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence $\frac{dz}{iz} = dx$ to obtain $$\frac{1}{2} \int_{|z|=1} \frac{1}{a+b(z-1/z)^2/4/(-1)}\frac{dz}{iz} \\ = \frac{1}{2} \int_{|z|=1} \frac{4}{4a-b(z-1/z)^2}\frac{dz}{iz} \\ = \frac{2}{i} \int_{|z|=1} ...


2

You can use integration by substitution $ 2x=sin(u) $. Then $ \frac{d}{du}x=\frac{1}{2}\cos(u) $. We can rearrange our substitution equation $ 2x=sin(u) $ into $ u = arcsin(2x) $. So we can find our limits with respect to u. When $ x = \frac{1}{8} \: \: u = arcsin(2\frac{1}{8}) = arcsin(\frac{1}{4}) $ and when $ x = 0 \: \: u = arcsin(0) = 0 $. So $ ...


0

Let $u=4x \Rightarrow du=4dx$ Therefore $\int\frac{4}{\sqrt{1-4x^2}}dx=8\int\frac{1}{\sqrt{4-u^2}}=8\arcsin(2x)+C$ $\Rightarrow\int_{0}^{1/8}\frac{4}{\sqrt{(1-4x^2)}}dx=8\arcsin(\frac{1}{4})$


2

$\int e^{kx}dx$ means that you need to find a lot of function which $(\ f + C \ )'= e^{kx}$. As you said $(e^{kx})'=k\cdot e^{kx} \Rightarrow (\frac{e^{kx}}{k} + C)'= e^{kx} \Rightarrow f=\frac{e^{kx}}{k}$ Therefore $\int e^{kx}dx=\frac{e^{kx}}{k}+C$, where C set of all constants.


2

Use the substitution $2x=\sin \theta$. Then $\frac{d}{d\theta}x=\frac{1}{2}\cos \theta $ and the integral becomes $$\int_{0}^{1/8} \frac{4}{\sqrt{(1-4x^2)}} \,dx = \int_{0}^{\arcsin\left(\frac{1}{4}\right)} 2\,d\theta$$


1

We find where the two surfaces intersect by setting the $z$'s equal to each other: $$x^2+3y^2=9-x^2$$ $$2x^2+3y^2=9$$ $$\frac{x^2}{\left(3/\sqrt 2\right)^2}+\frac{y^2}{\left(\sqrt 3\right)^2}=1$$ So the projection of the intersection onto the $xy$ plane is an ellipse centered at the origin. You should know several ways to express that bounds of that region ...


1

The second function does not represent the graph of a cylinder, rather an open parabola (graph it!). Now, the first function is a paraboloid---a deformed "bowl". You want the volume under the open parabola and inside the bowl. So: 1). Integrate from $z=x^{2}+3y^{2}$ to $z=9-x^{2}$. Now you have $\int \int \int_{x^{2}+3y^{2}}^{9-x^{2}}dzdydx$. 2). Next, ...


1

you have $$ydy = \frac{\tan^{-1}x}{1 + x^2}dx, \quad y = -1, x = 1 \tag 1$$ with a change of variable $u = \tan^{-1} x , du = \frac{dx}{1+x^2} x = 1\to u = \pi/4,$ $(1)$ becomes $$\int_{-1}^yydy = \int_{\pi/4}^uudu \to y^2 - 1 = u^2 - \pi/4, y = -\sqrt{u^2 - \pi/4 + 1} $$ therefore $$y = -\sqrt{\tan^2 x - \pi/4 + 1} $$



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