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1

You did everything right, but you're constant of integration "absorbs" the $\frac{1}{24}$ $$\frac{1}{24}+C=C$$ The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant! If it helps, you could do something like: ...


2

You missed the constant of integration all the way! It "absorbs" the $\frac{1}{24}$! Spot the difference: Evaluating: $\displaystyle \frac{1}{6} \cdot \frac{s^2}{2} +c= \frac{s^2}{12} +c= \frac{\cos^2(u)}{12} +c= \frac{\cos^2(\frac{6}{x})}{12}+c$, where c is the constant of integration. Then using a half angle formula for $\cos^2$: $\displaystyle \frac ...


0

Setting $$ f(x)=\int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy, \quad x\in (-1,1), $$ we have $f(0)=0$, and $$ f'(x)=\int_0^\pi\frac{\cos y}{(1+x\cos y)\cos y}\,dy=\int_0^\pi\frac{1}{1+x\cos y}\,dy\quad \forall x\in (-1,1) $$ Setting $$ t=\tan\frac{y}{2}, $$ we have $$ y=2\arctan t,\, \frac{1}{1+x\cos ...


2

Just to make things easier on my eyes (I hate all of the 11's) i'd start off with: $let\:v = 11x,\: dv = 11dx$ $$\frac{1}{11}\int{\frac{\sec v \:\tan v}{\sqrt{\sec v}}}\:dv$$ $let\: u=\sec v,\: du = \sec v\:\tan v\:dv$ $$\frac{1}{11}\int{\frac{1}{\sqrt{u}}}\:du$$ You can take it from here.


4

Your $du$ is wrong. It should have an extra 11 in it. :)


1

$\bf{My\; Solution}$ We Know that in $$x\in (0,1)\;\;, x^2<x\Rightarrow -x^2>-x\Rightarrow e^{-x^2}>e^{-x}$$ So $$\displaystyle \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1}e^{-x}dx = \left(1-\frac{1}{e}\right)$$ and in $$\displaystyle x\in (0,1)\;, x^2>0\Rightarrow e^{-x^2}<e^{-0}$$ So $$\displaystyle ...


0

Probably the most direct way to do this is by infinite series: expanding the logarithm, and interchanging the sum and integral, we have $$ I := \int_0^{\pi} \frac{\log{(1+x\cos{y})}}{\cos{y}} \, dy = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} \int_0^{\pi} \cos^{n-1}{y} \, dy. $$ Firstly, note that $n-1$ has to be even for the integral to be nonzero, since ...


1

The answer is both. Differentiate and you will get the original integrand in both cases. Why? $\frac{1}{2}\ln{(\frac{5}{2} + e^x)} = \frac{1}{2}\ln{\frac{5+ 2e^x}{2}} = \frac{1}{2}(\ln{(5+ 2e^x)} - \ln{2})$, which differs from the other answer by a constant. Therefore, both antiderivatives are correct.


3

$$C + \frac 12 \ln(5+2e^x ) =C + \frac 12\ln[2(\frac 52 +e^x)]=C +\frac12\ln 2 + \frac 12 \ln\left(\frac 52 + e^x\right) $$ the two antiderivatives differ by a constant.


0

Since $$\frac{\mathsf d}{\mathsf dx}\left[\frac12\log(5+2e^x)\right] = \frac{2e^x}{2(5+2e^x)}=\frac{e^x}{5+2e^x}, $$ it follows that $\frac12\log(5+2e^x)$ is an antiderivative of $\frac{e^x}{5+2e^x}.$


1

Call $u = 5+2e^x$ right off the bat. So ${\rm d}u = 2e^x\,{\rm d}x$, and: $$\int \frac{e^x}{5+2e^x}\,{\rm d}x = \int\frac{1}{2u}\,{\rm d}u = \frac{1}{2}\ln u + c = \frac{1}{2}\ln(5+2e^x)+ c, \quad c \in \Bbb R.$$


2

Your expression for the Taylor series is wrong, it should be $$f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \ldots$$ What you've written isn't a power series, since each term after the second are constants rather than constants times powers of $x$. (Also, you should be careful to not ...


0

I think a different substitution's probably the way forward: try instead $x=n(1+u)$, so $dx = n \, du$. Then if the contents of the limit is $I_n$, $$ I_n = \int_0^{\infty} \frac{n^{\alpha}}{(n(1+u))^{\alpha+1}} f\left( \frac{1+u}{n} \right) (nu) n \, du, $$ which cancels down to $$ I_n = \int_0^{\infty} \frac{u}{(1+u)^{\alpha}} \left[ f\left( \frac{1+u}{n} ...


0

Well from the Pythagorean Identities of trig we have: $\tan^{2}(\theta) + 1 = \sec^{2}(\theta)$ this tells us that if we let $x=a\sec\theta$ we can then reduce this to solving $$\int\frac{a\sec\theta\tan\theta}{a^2\sec^2\theta-a^2}d\theta$$ I'm assuming you can take it from here.


3

Note that $\frac{x^n}{x^n+1}=1-\frac{1}{x^n+1}$. Thus, $$\int_1^2 \frac{x^n}{x^n+1}dx=1-\int_1^2 \frac{1}{1+x^n}dx$$ For $n>1$, the integral on the right-hand side satisfies the inequality $$\left|\int_1^2 \frac{1}{1+x^n}dx\right|\le \int_1^2 x^{-n}dx=\frac{1-2^{1-n}}{n-1}$$ which clearly goes to zero as $n \to \infty$. Putting all of this together ...


0

Hint: Use $\log(1+x)=-\sum_{k=1}^{\infty}\frac{(-x)^k}{k}$ ,then integrate termwise.


5

Ok, just a hint: $$\int \frac{x^2}{x^2+1}\,{\rm d}x = \int 1 - \frac{1}{1+x^2}\,{\rm d}x,$$ and the last one is immediate.


4

Hint: $$x^2 = x^2 + 1 - 1 $$ or you could try this substitution: $$x = \tan(\theta)$$


0

Divide the numerator and denominator of the integrand by $x^n$: $$ \frac{1}{1+x^{-n}} $$ For every $x \in (1,2]$, $x^{-n} \to 0$ as $n \to \infty$. Since this is all but one point of the interval of integration, the integral tends to $$ \int_1^2 \frac{1}{1} \, dx = 1. $$ To do it more thoroughly than this, you can chop up the interval into a small region ...


1

here we go. what you have is $$T(x,y) = F(x+y) - F(x-y) \text{ where } F'(t) = \ \frac{\sin t}t.$$ now we can differentiate $T$ with respect to $x$ to get $$T_x = F'(x+y) \times 1 - F'(x-y) \times 1 =F'(\pi) - F'(0))= 1 -0= 1.$$


6

Well, $\frac{x^n}{x^n+1} \leq 1$ for all $x \in [1,2]$, and the constant function $1$ is integrable in $[1,2]$, so by the Dominated Convergence Theorem we have: $$\lim_{n \to +\infty} \int_1^2\frac{x^n}{x^n+1}\,{\rm d}x = \int_1^2 \lim_{n \to +\infty} \frac{x^n}{x^n+1}\,{\rm d}x = \int_{1}^2 1\,{\rm d}x = 1.$$


1

$$\frac{\partial{T}}{\partial{x}}=\frac{\sin(x+y)}{x+y}-\frac{\sin(x-y)}{x-y}$$ Replacing $x=\frac{\pi}{2}$ and $y=-\frac{\pi}{2}$ we get $$T_x(\frac{\pi}{2},-\frac{\pi}{2})=1$$ Using the fact that $\lim\limits_{x\to 0}\frac{\sin{x}}{x}=1$ and $\sin{\pi}=0$ I used the chain rule to get the initial derivative but it is trivial because $(x+y)_x=(x-y)_x=1$


1

We can think of: $$T(x,y) = \int_{a(x,y)}^{b(x,y)}f(t)\,{\rm d}t,$$ with $a(x,y) = x-y$, $b(x,y) = x+y$ and $f(t) = \sin(t)/t$. We have: $$T(x,y) = \int_0^{b(x,y)} f(t)\,{\rm d}t -\int_0^{a(x,y)}f(t)\,{\rm d}t, $$ so: $$\frac{\partial T}{\partial x}(x,y) = \frac{\partial b}{\partial x}(x,y) f(b(x,y)) - \frac{\partial a}{\partial x}(x,y) f(a(x,y)).$$ Now ...


2

You may just perform the change of variable $u=2+x^{9/2}$, $du=\dfrac92x^{7/2}\:dx$, giving $$ \int x^{7/2} \sec^2(2+x^{9/2})\: \mathrm{d}x=\dfrac29\int \sec^2(u) \:\mathrm{d}u= \dfrac29\tan (u)+C. $$ Can you take it from here?


3

Substitute $u=$ the term inside the parentheses.


2

You solved the ODE correctly. We have $$z(t) = -\ln(-4t^2+c) \implies z(0) = -\ln(c) \implies \ln(c) = 0 \implies c = 1,$$ because $\ln 1 = 0.$ Ok?


0

you can solve this O.D.E as follows $$\frac{dB}{dx}+2B=0$$ $$m+2=0$$ $$m=-2$$ $$y_c=C_1e^{-2x}$$ to find the particular solution $$y_p=A$$ then the $A=25$ hence $$y=C_1e^{-2x}+25$$


3

Any simple function $0\leq\phi\leq f$ satisfies $0\leq\phi\leq g$, hence $$\int_Xfd\mu=\sup_{0\leq\phi\leq f,\ \phi\ simple} \sum_{k\in\mathbb{R}}k\cdot\mu(\phi^{-1}(k))\leq\\\leq\sup_{0\leq\phi\leq g,\ \phi\ simple} \sum_{k\in\mathbb{R}}k\cdot\mu(\phi^{-1}(k))\leq\int_Xgd\mu$$ $g(x)\leq0\rightarrow f(x)\leq g(x)\leq0$, hence $f^-\geq g^-$. Moreover, ...


1

separating the variable works: $$ \frac {dB}{dx} = 50 - 2B \\ \int \frac{dB}{25 - B} = \int 2dx + C, C\in\Bbb R \\ -\log |25 - B| = 2x + C, C\in\Bbb R $$ yields the general solution: $$ B = 25 + K\exp (-2x), K\in\Bbb R $$ and with the initial condition: $$ B = 25 (1+ \exp (2(1-x))) $$


0

$$\int \cosh^3(x)dx=\int (\frac{e^x+e^{-x}}{2})^3dx=\frac{1}{8}\int (e^{3x}+3e^{x}+3e^{-x}+e^{-3x})dx$$ $$=\frac{1}{8} (\frac{1}{3}e^{3x}+3e^{x}-3e^{-x}-\frac{1}{3}e^{-3x})+C$$ $$=\frac{2}{24}\sinh(3x)+\frac{3}{4}\sinh(x)+C$$


1

Hint this is better to integrate $$\frac{e^{-3 x}}{8}+\frac{3 e^{-x}}{8}+\frac{3 e^x}{8}+\frac{e^{3 x}}{8}$$ integrating this we get $$-\frac{e^{-3x}}{24}-\frac{3}{8}e^{-x}+\frac{3}{8}e^{x}+\frac{e^{3x}}{24}+C$$


0

Are you sure about what is $<f,f>$? Try an example, let say $f(x)=\cos x$. Observe that $$<f,f>=\int_0^1 [f(x)]^2\, dx$$ and $[f(x)]^2\geqslant 0$ for all $x$. Now you can use the fact that "the integral is the area under the curve" to argue that $<f,f>$ is always nonnegative.


6

Hint: $\cosh^3x=\cosh x\cdot\cosh^2x=\sinh'x\cdot\big(1+\sinh^2x\big)$.


0

The exponential term can be slightly rearranges as $$e^{-(1/\Delta t-i(E_0-E)/h)t}=e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}$$ Taking the magnitude of the right-hand side and exploiting the fact that for real-valued $x$, $|e^{ix}|=|\cos x + i \sin x|=\sqrt{\cos^2x+\sin^2x}=1$, we find $$0\le |e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}|\le e^{-(1/\Delta t)t}$$ and the ...


0

The idea is to use the identity $$ \cosh^2 u = \sinh^2 u + 1 $$ to simplify the square root; so, let $$ \sinh^2 u = 4y^2 \to \sinh u = 2y \to u = \sinh^{-1} 2y \\ \implies \cosh u du = 2dy \\ \sqrt {1 + 4y^2} = \cosh u \\ $$ from this you should be able to reoslve the integral.


2

It's a matter of principle.[*] A $k$-form eats $k$ vectors and spits out a number. If you want to integrate a $k$-form over a $m$-dimensional submanifold, how do you choose which $k$ of the tangent vectors you're going to plug in to the form on each tangent space? If $k<m$, you'll have leftover vectors and if $k>m$ you won't have enough vectors. Either ...


8

$$\begin{aligned} I_{n-1}(r)+I_{n+1}(r) &=\int_0^{\pi}\frac{2\cos(nx) \cos x}{r^2-2r\cos x+1}\,dx \\ &=-\frac{1}{r}\int_0^{\pi} \frac{\cos(nx)(r^2-2r\cos x+1-r^2-1)}{r^2-2r\cos x+1}\,dx \\ &=-\frac{1}{r}\int_0^{\pi}\cos(nx)\,dx+\frac{r^2+1}{r}\int_0^{\pi} \frac{\cos nx}{r^2-2r\cos x+1}\,dx\\ &=\left(r+\frac{1}{r}\right)I_n(r) \\ ...


0

Trough a lot of trouble I got: $$\int \frac{x}{\sqrt{x^2-6x}}dx=\frac{x(x-6)+6\sqrt{x-6}\sqrt{x}*ln\left(\sqrt{x-6}\sqrt{x}\right)}{\sqrt{(x-6)x}}+C$$


-3

$$\int_0^\pi \frac{x}{(\sin x)^{\sin (\cos x)}}dx=$$ $$x\int_0^\pi sin^{-sin(cos(x))}(x) dx$$ I got no result in terms of standard mathematical functions!


0

Every continuous function on $\mathbb R$ has an antiderivative, by the version of the fundamental theorem of calculus that says that if $f$ is continuous on $\mathbb R$, then the function $g(x)=\int_0^x f(t)\,dt$ is an antiderivative of $f$. Therefore $\sin^n(x)$ has an antiderivative on $\mathbb R$. By definition, every antiderivative is differentiable. ...


1

The first series has the value $$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\left(\psi{\left(\frac{n}{2}\right)}+\frac{1}{n}\right)^2-\left(\psi{\left(\frac{n+1}{2}\right)}\right)^2}{n}=\frac83\ln^3{(2)}-\frac34\,\zeta{(3)}+2\gamma\,\ln^2{(2)}.$$ A useful formula here is the following addition identity for the digamma function: ...


1

Since the vector field $\langle x_2^2 \cos(x_1), 2x_2(1 + \sin(x_1))\rangle$ is the gradient of the scalar field $\phi(x_1,x_2) = x_2^2(1 + \sin(x_1))$, by the fundamental theorem of line integrals, your integral is zero.


-1

Hint: if you have $f(x)=F'(x)$, you can change the ''symbol'' for the variable, i.e write $t=x$, and you have: $f(t)=F'(t)$. Than: $$ \int_a^b f(x)=F(x)|_a^b=F(b)-F(a) $$ $$ \int_a^b f(t)=F(t)|_a^b=F(b)-F(a) $$ In your case you have: $$ \int_0^\pi\dfrac{\sin (t+\pi)}{t+\pi}dt=-\int_0^\pi\dfrac{\sin t}{t+\pi}dt=-\int_0^\pi\dfrac{\sin x}{x+\pi}dt $$


0

Stop torturing yourself with parametrizations and use Green's theorem: you'll get $0$ in just a few seconds.


2

Put $u = x^2+4 \implies du = 2x\,dx\implies x\,dx = \frac 12 du$. $$ \begin{align}\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x &= \frac 12\int u^{-5} \,du\\ & = \frac 12 \left(\frac {u^{-4}}{-4}\right) +c \\ &= -\frac{1}{8u^4} + c \\ &= -\frac {1}{8(x^2 + 4)^4} + c\end{align} $$


1

with $u = x^2 + 4$: $$ \int (x^2 + 4)^{-5 } {x dx} = \frac 12 \int u^{-5} {du} = - \frac 1{8} u^{-4} = - \frac 1{8} (x^2 + 4)^{-4} $$


0

It is pretty easy to integrate functions of the form $\sin(mx)$ or $\cos(mx)$ for $m\in\mathbb{Z}$, so we just need to show that for every $n\in\mathbb{N}$ the function $\sin^n(x)$ can be expressed as a linear combination of them, i.e. consider the Fourier series. Assuming $n=2k$ we have: $$ \sin^n(x) = ...


3

I don't know what you mean by "transfer" but: Hint Note that $d(x^2 + 4) = 2 x \,dx$, so the $x$ in the numerator suggests a particular substitution.


2

You may use Weierstrass approximation theorem, providing a polynomial $p(s)$ for which: $$\forall s\in[0,1],\quad \left|g(s)-p(s)\right|\leq \varepsilon,\quad p'(0)=0, \tag{1}$$ then, for every $t\in[0,1)$, $$(1-t)\left|\int_{0}^{t}\frac{g(s)}{(1-s)^2}\,ds-\int_{0}^{t}\frac{p(s)}{(1-s)^2}\,ds\right|\leq\varepsilon \tag{2}$$ and now we are allowed to use ...



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