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12

let $x=\cos u$ so that $dx=-\sin u \, du$ \begin{align} & =\int_{\pi /2}^{0}\sqrt{\frac{1-\cos u}{1+\cos u}}(-\sin u) \, du \\ & =\int_{0}^{\pi /2}\tan (u/2)\sin u \, du \\ & =\int_{0}^{\pi/2}\frac{\sin u}{1+\cos u}\sin u \, du=\int_0^{\pi/2} \frac{1-\cos^2 u}{1+\cos u} \, du \\ & =\int_{0}^{\pi/2}(1-\cos u) \, ...

7

Multiply both numerator and denominator by $\sqrt{1-x}$, then $\sqrt{\frac{1-x}{1+x}}$ becomes $\frac{1-x}{\sqrt{1-x^2}}$. The integral of first term is $\arcsin x$, the second term can be integrated by substitution.

5

Let the integral be $I$. Do the substitution $y=2016-x$. This leads to the integral $$I = \int_2^{2014} \frac{\log{(y-1)}}{\log{(y-1)}+\log{(2015-y)}} \, dy,$$ and you can relabel $y$ as $x$ and add together the two ways of writing $I$ to obtain $$2I = \int_2^{2014} \frac{\log{(2015-x)}+\log{(x-1)}}{\log{(x-1)}+\log{(2015-x)}} \, dx = \int_2^{2014} 1 \, ... 4 Andrey Kaipov gave the nice answer to the question. You can also use approximation using Taylor series of Taylor series around x=0 and arrive to$$\sin\big(\cos(x)\big)=\sin (1)-\frac{1}{2} x^2 \cos (1)+x^4 \left(\frac{\cos (1)}{24}-\frac{\sin (1)}{8}\right)+x^6 \left(\frac{\sin (1)}{48}+\frac{7 \cos (1)}{360}\right)+x^8 \left(\frac{\sin ...

3

As the value of the cosine remains in $(-1,1)$, you can replace the sine by its Taylor development, and get reasonable convergence speed. $$\sin(\cos(x))=\sum_{k=0}^\infty\frac{(-1)^k\cos^{2k+1}(x)}{(2k+1)!}$$ The area under the odd powers of the cosine is well-known $$\int_0^{\pi/2}\cos^{2k+1}(x)\,dx=\frac{(2k)!!}{(2k+1)!!}.$$ After simplification, ...

3

Heuristically, $\sin x \approx. x$ for small $x$. Thus, $$\int_{1/(n+1)}^{1/n} \frac{\sin x }{x^3}dx\approx. \int_{1/(n+1)}^{1/n} \left(\frac{1}{x^2} \right)dx\ =1$$ We can make this argument rigorous by writing $$\frac{\sin x}{x^3}=x^{-2}+\sum_{k=1}^{\infty} \frac{x^{2k-2}}{(2k+1)!}$$ whereupon $$\int_{1/(n+1)}^{1/n} \frac{\sin x ... 2 You may find this integral at DLMF : Modified Bessel function 10.32.10\; (setting z=2\,\sqrt{x},\; \nu=0) :$$K_{\nu}(z)=\frac 12\left(\frac z2\right)^{\nu}\int_0^\infty e^{\large{-t-\frac {z^2}{4t}}}\frac {dt}{t^{\nu+1}}$$so that you should obtain \;I(x)=2\;K_0(2\sqrt{x})\, as indicated by mickep. You may rewrite it too as 10.32.6 : ... 2 Hint: Let x = a \sin t  then dx = a \cos t\, dt and$$\int_{-\pi/2}^{\pi/2} \sqrt{a^2 - a^2 \sin^2 t}\,\, a\cos t \, dt =\int_{- \pi/2}^{ \pi/2} a^2\cos^2 t \, dt$$2 If y=\sqrt{a^2-x^2} then y \geq 0 and$$x^2+y^2=a^2$$This shows that your integral represents the area between the x-axis and the upper half circle x^2+y^2=a^2. The area of this half disk is \frac{\pi a^2}{2}. 2 Okay so the integral can be evaluated to be$$ \begin{align*} \int \frac{e^{x/2}}{2}+ \frac{1}{2x}− \frac{3}{2} \; dx = e^{x/2} + \frac{\ln{x}}{2} - \frac{3x}{2} + c = c \end{align*} $$and the derivative$$ \frac{d}{dx}\left(\frac{e^{x/2}}{2}+ \frac{1}{2x}− \frac{3}{2}\right) = \frac{e^{x/2}}{4} - \frac{1}{2x^2} = 0 $$2 There is no elementary antiderivative, but the integral is the Struve function, \text{H}_\alpha(x), for \alpha=0, and x=1. Then by definition and Wolfram Alpha's calculations,$$\int\limits_{-\pi/2}^{\pi/2}\sin\cos(x)dx = 2\int\limits_{0}^{\pi/2}\sin\cos(x)dx = \pi\ \text{H}_0(1) \approx1.786.$$2$$ \begin{align} \int_0^1\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x &=2\int_{1/2}^1\sqrt{\frac{1-x}{x}}\,\mathrm{d}x\tag{1}\\ &=4\int_{1/\sqrt2}^1\sqrt{1-x^2}\,\mathrm{d}x\tag{2}\\ &=4\int_0^{\pi/4}\sin^2(x)\,\mathrm{d}x\tag{3}\\ &=2\int_0^{\pi/4}1-\cos(2x)\,\mathrm{d}x\tag{4}\\ &=\left[2x-\sin(2x)\vphantom{\int}\right]_0^{\pi/4}\tag{5}\\[6pt] ...

2

Take $f(\frac{1}{2},y)$ to be the Dirichlet function in $y$ and $f(x,y) = 1$ otherwise. $f$ is Riemann integrable on $[0,1] \times [0,1]$ but $g(y) = f(\frac{1}{2},y)$ is not Riemann integrable on $[0,1]$. $f$ is Riemann integrable on the unit square because the partition of $x$ containing $x = \frac{1}{2}$ can be made arbitrarily small. Since the infimum ...

2

This solid is not obtained by a rotation. Here, the area of the triangle $T_y$ that cross the y axis at $y$ is given by $A(y) = \frac{1}{2}(2x)^2 = 2(1-y^2)$, because $x^2+y^2=1$ Hence the volume of your solid is $$\int_{-1}^1 A(y) dy = \int_{-1}^1 2(1-y^2)dy = \frac{8}{3}$$

2

the characteristic feature of a logarithmic (or 'equi-angular') spiral is central self-similarity, expressed geometrically as a proportionality between the two elements of length in polar cordinates. we may write this: $$dr = \alpha r d\theta \tag{1}$$ which gives, for the element of length $dl$: $$dl = \left(dr^2 + (rd\theta)^2) \right)^{\frac12}=\beta ... 1 In the end we don't want the gradient term on the RHS of the inequality:$$\int_U|Du|^p\,dx\le C\left(\int_U |u|^\frac{p}{2} |D^2u|^\frac{p}{2}\right)^\frac{2}{p} \left(\int_U |Du|^p \right)^\frac{p-2}{p},$$dividing by \left(\int_U |Du|^p \right)^\frac{p-2}{p}, we obtain$$\left(\int_U |Du|^p \right)^{\frac{2}{p}}\le C\left(\int_U |u|^\frac{p}{2} ...

1

The integral for calculating the area "under" the $y$-axis involves two components. Note that for $1\le y\le 3$, the function $x(y)=\pm \sqrt{3-y}$ is double-valued. Thus, $$\text{Area} = \int_{1}^{3} \left(\sqrt{3-y}-(-\sqrt{3-y})\right) \,dy=2\int_{1}^{3} \sqrt{3-y} \,\, dy$$ which will recover the correct answer $\frac{32}{3}$.

1

Selecting two points $x$ and $y$ from the interval $[0, 1]$ is identical to selecting a single point from the unit square. For the sake of clarity, let us calculate the probability that $y/x$ is closest to an even integer, because that's just the slope. By symmetry, that is equal to the corresponding probability for the ratio $x/y$. Divide the unit square ...

1

Hint: the area of each of these triangles is half the height of a circle squared at the given point. Hint #2: The height of the circle is going to be $2\sqrt{1-x^2}$, right? But we can figure out the area of the triangle only from the base!

1

We have $$\dfrac{(x^2+x+1)^n-1}{x^2+1} = \dfrac{\displaystyle\sum_{k=0}^n \dbinom{n}k x^{n-k}(x^2+1)^k-1}{x^2+1} = \sum_{k=1}^n \dbinom{n}k x^{n-k} (x^2+1)^{k-1} + \dfrac{x^n-1}{x^2+1}$$ Clearly, we have $\displaystyle\int_0^1 x^{n-k}(x^2+1)^{k-1}dx \in \mathbb{Q}$ for all $n \geq k \geq 1$. Hence, $$\displaystyle \sum_{k=1}^n \dbinom{n}k \int_0^1 x^{n-k} ... 1 Hint: Use this substitution.$$x = a\sin(\theta)$$1 We have$$\int\frac{x^{3}e^{x^{2}}}{x^{2}+1}dx=\int\left(xe^{x^{2}}-\frac{xe^{x^{2}}}{x^{2}+1}\right)dx=\frac{1}{2}e^{x^{2}}-\frac{\textrm{Ei}\left(x^{2}+1\right)}{2e}+C $$where \textrm{Ei}\left(x^{2}+1\right)  is the exponential integral. Maybe it's useful to remaind that$$\frac{d}{dx}\textrm{Ei}\left(x\right)=\frac{e^{x}}{x}. $$1 Your idea "the function is continuous hence the anti-derivative exits" doesn't work. This is true for real functions, but it is not true anymore in the complex case. For example, complex conjugation z\mapsto \bar z is continuous, but is not the derivative of anything (on any nonempty open set). After a course in complex analysis, the argument that ... 1 HINT: Maybe using this simple inequality and then squeeze it?$$x-\frac{x^3}{6}\le \sin(x) \le x \ , x\ge0$$The limit is 1. 1 Let's generalize. Why? Because this kind of limit doesn't really depend on power series or anything fancy. Suppose we have a continuous f on (0,1) with \lim_{x\to 0^+}f(x) = 1. Then$$\int_{1/(n+1)}^{1/n}\frac{f(x)}{x^2}\,dx \to 1$$as n \to \infty. (In our problem we have f(x) = (\sin x)/x.) Proof: Let \epsilon>0. Choose \delta > 0 ... 1 Hint: \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots 1 You did a taylor expansion for g around x_0, neglected all terms of order 3 or above. Since x_0 is a maximum, you have that g'(x_0)=0 and g''(x_0)\leq 0, so$$g(x)\approx g(x_0)+\frac{1}{2}(x-x_0)^2g''(x_0)=g(x_0)-\frac{1}{2}(x-x_0)^2|g''(x_0)|. You used that $g''(x_0)\leq0$ to write $g''(x_0)=-|g''(x_0)|$.

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