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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: ...
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$$You can take it from here. 2 du = sec(11x) tan (11x) (11dx) 4 Your du is wrong. It should have an extra 11 in it. :) 1 \bf{My\; Solution} We Know that in$$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=0m+2=0m=-2y_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$u = x^2+4 \implies du = 2x\,dx\implies x\,dx = \frac 12 du. \begin{align}\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x &= \frac 12\int u^{-5} \,du\\ & = \frac 12 \left(\frac {u^{-4}}{-4}\right) +c \\ &= -\frac{1}{8u^4} + c \\ &= -\frac {1}{8(x^2 + 4)^4} + c\end{align} 1 withu = x^2 + 4$: $$\int (x^2 + 4)^{-5 } {x dx} = \frac 12 \int u^{-5} {du} = - \frac 1{8} u^{-4} = - \frac 1{8} (x^2 + 4)^{-4}$$ 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 ...