# Tag Info

13

There are a couple of ways to see this. Firstly, draw a right triangle, call it $ABC$ (with $C$ being the right angle), with side lengths $a$, $b$ and $c$ with the usual convention. Then $\arcsin(\frac{b}{c})$ is the measure of the angle $CBA$. Additionally, $\arccos(\frac{b}{c})$ is the angle of the angle of the opposite angle $CAB$, so ...

8

In this case it's exactly the same: \begin{align} \left|\frac{3+\sqrt{9-x^2}}{x}\right|^{\!-1} &= \left|\frac{x}{3+\sqrt{9-x^2}}\right|\\[2ex] &= \left|\frac{x}{3+\sqrt{9-x^2}}\frac{3-\sqrt{9-x^2}}{3-\sqrt{9-x^2}}\right|\\[2ex] &=\left|\frac{x(3-\sqrt{9-x^2})}{9-9+x^2}\right|\\[2ex] &=\left|\frac{3-\sqrt{9-x^2}}{x}\right|\\[2ex] ...

6

Let $\{x_0,x_1,\dots,x_N\}$ be a partition of $[a,b]$. Then $\{f(x_0),f(x_1),\dots,f(x_N)\}$ is a partition of $[f(a),f(b)]$. The following equality holds: $$\sum_{i=0}^{N-1}f(x_i)(x_{i+1}-x_i)+\sum_{i=0}^{N-1}x_i(f(x_{i+1})-f(x_i))+\sum_{i=0}^{N-1}(x_{i+1}-x_i)(f(x_{i+1})-f(x_i))=b\,f(b)-a\,f(a).$$ The first two sums are Riemann sums for $\int_a^bf$ and ...

6

Here is a hint only: Write the sum as $$\sum_{i=1}^n \frac{1}{\bigl(\frac{i-1}{n}\bigr)^2+1}\cdot \frac{1}{n}$$ Do you recognize it better now as a Riemann sum?

5

$$\lim _{n\rightarrow \infty }\sum_{i=1}^{n}\frac{n}{\left ( i-1 \right )^{2}+n^{2}}=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{1+\left(\frac{k}{n}\right)^2}=\int_{0}^{1}\frac{dx}{1+x^2}=\arctan 1=\color{red}{\frac{\pi}{4}}.$$

5

The integral as stated does not converge. On the other hand, its Cauchy principal value exists and may be computed using the residue theorem. Consider the integral $$\oint_C dz \frac{\sqrt{z}}{z^2-1}$$ where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$, with semicircular detours of radius $\epsilon$ into the upper half plane ...

5

A way to compute this is as follows: \begin{align*} \int_0^1 \frac {x^3}{\sqrt {4+x^2}}\mathrm d x &=\int_0^1\frac{4x+x^3-4x}{\sqrt{4+x^2}}\mathrm d x\\ &=\int_0^1x\sqrt{4+x^2}\mathrm d x -2\int_0^1\frac{2x}{\sqrt{4+x^2}}\mathrm d x\\ &=\left.\frac{1}{3}(4+x^2)^{3/2}\right|_0^1-4\left.\left(4+x^2\right)^{1/2}\right|_0^1\\ ...

5

I think it is very natural from a geometrical point of view. It's just about the addition of two areas, which make up a big rectangle substracting a small one. See the graph below: Now, obviously, in the case shown in my graph $$S_1=\int_{a}^{b}f(x)dx$$ and $$S_2=\int_{f(a)}^{f(b)} f^{-1}(x)dx$$ Geometrically, we have $$S_1+S_2=S_{big}-S_{small}$$ where ...

5

The answer in your textbook is wrong. $x\mapsto\tan x$ is an odd function, hence $$\int_{-a}^{a}\tan\theta\,\mathrm d\theta=0,$$ for all $a\in\left(-\tfrac\pi2,\tfrac\pi2\right)$.

5

Multiply numerator and denominator by $1+\cos x$ to get \begin{align}\int_{0.5\pi}^\pi \frac{(1+\cos x)\,dx}{1-\cos^2 x}& = \int_{0.5\pi}^\pi \frac{(1+\cos x)\,dx}{\sin^2 x} \\ \\ &= \int \csc^2 x \,dx + \int \frac{\cos x\,dx}{\sin^2 x}\\ \\ & = -\cot x + \int \cot x\csc x\,dx\\ \\ &=-\cot x -\csc x+C\end{align}

4

Hint Let us look at the antiderivative and, first, make a change of variable $1+ay=x$. So, $$I=\int\frac{e^{-cy}}{1+ay}\,dy=\frac{e^{c/a}}{a}\int\frac{e^{-cx/a}}{x}\,dx$$ Now, change again setting $\frac{cx}{a}=z$; you then arrive to $$I=\frac{e^{c/a}}{a}\int\frac{e^{-z}}{z}\,dz=\frac{e^{c/a}}{a}\text{Ei}(-z)$$ where appears the exponential integral.

4

First Approach If $y=mx$ and $x^2+y^2=\frac1m$, then $x^2+y^2=\frac xy$, which implies $$yx^2-x+y^3=0\tag{1}$$ Taking the implicit derivative of $(1)$: $$y'=\frac{1-2xy}{x^2+3y^2}\tag{2}$$ $y'=0$ means $1=2xy$; then applying $y=mx$, we get $$1=2xy=2mx^2\tag{3}$$ applying $x^2+y^2=\frac1m$ gives $$1+2my^2=2mx^2+2my^2=2m\tag{4}$$ using $y=mx$ yields ...

4

It is pretty obvious to see that $\sin$ and $\cos$ are the same curve, just shifted by $\pi/2$, so if you consider following craphis, it should be clear:

4

All integrals of this form are solved by letting $t=\dfrac1{1+x^b}$ and then recognizing the expression of the beta function in the new integral.

4

Let $\delta_1>0,\delta_2<1$ real numbers close to zero and one, respectively, and I=$(\delta_1,\delta_2)\cup(\delta_2^{-1},\delta_1^{-1})$. We have: $$\int_{I}\frac{\sqrt{x}}{x^2-1}\,dx = \int_{\delta_1}^{\delta_2}\frac{\sqrt{x}}{x^2-1}\,dx +\int_{\delta_1}^{\delta_2}\frac{\sqrt{1/x}}{1-x^2}\,dx = ... 3 By changing the variable t=\frac\pi2-x the integral becomes$$\int_0^{\pi/2}\frac{t^\alpha}{(\sin t)^\beta}dt$$and since$$\frac{t^\alpha}{(\sin t)^\beta}\sim_0\frac1{t^{\beta-\alpha}}$$so the given integral is convergent if and only if \beta-\alpha<1. 3$$\frac{e^{-x^2}-e^{-x}}{x} = \int_x^1 dt \, e^{-x t} = \int_{x^2}^x dy \, e^{-y}$$Therefore, let's integrate and reverse the order of integration as follows.$$\begin{align} \int_0^{\infty} dx\, \frac{e^{-x^2}-e^{-x}}{x} &= \int_0^{\infty} \frac{dx}{x} \, \int_{x^2}^x dy \, e^{-y} \\ &= \int_0^{\infty} dy \, e^{-y} \int_y^{\sqrt{y}} \frac{dx}{x} ...

3

By definition, $\arcsin(x)$ is the angle $\alpha$ such that $\sin(\alpha) = x$ and $-\pi/2 \le \alpha \le \pi/2$, while $\arccos(x)$ is the angle $\beta$ such that $\cos(\beta) = x$ and $0 \le \beta \le \pi$. Since $-\pi/2 \le \alpha \le \pi/2$, $\cos(\alpha) \ge 0$, so we have $\cos(\alpha) = \sqrt{1 - x^2}$. Similarly $\sin(\beta) = \sqrt{1-x^2}$. Now ...

3

\begin{align} \int_{-\infty}^{\infty}x\mathrm{e}^{-|x|}dx & = \int_{-\infty}^{0}x\mathrm{e}^{x}dx + \int^{\infty}_{0}x\mathrm{e}^{-x}dx \\[6pt] & =\int_{\infty}^{0}x\mathrm{e}^{-x}dx + \int^{\infty}_{0}x\mathrm{e}^{-x}dx \\[6pt] & =-\int^{\infty}_{0}x\mathrm{e}^{-x}dx + \int^{\infty}_{0}x\mathrm{e}^{-x}dx \\[6pt] & =0 \end{align}

3

\begin{align} \int_0^\infty x \Big( e^{-x}\,dx\Big) & = \int x\,dv = xv-\int v\,dx \\[8pt] & = \left.-xe^{-x}\vphantom{\frac11}\right|_0^\infty - \int_0^\infty -e^{-x}\,dx \\[8pt] & = \left.\left(-xe^{-x} - e^{-x}\right)\vphantom{\frac11}\right|_0^\infty. \end{align} It is easy to evaluate this expression at $x=0$. To "evaluate it at $\infty$" ...

3

Let $C$ be the graph of $y = f(x)$ over the interval $[a,b]$. Then $\int_a^b f(x)\, dx$ is the line integral $\int_C y\, dx$, and $\int_{f(a)}^{f(b)} f^{-1}(y)\, dy$ is the line integral $\int_C x\, dy$. Thus $$\int_a^b f(x)\, dx + \int_{f(a)}^{f(b)} f^{-1}(y)\, dy = \int_C x\, dy + y\, dx = \int_C d(xy) = bf(b) - af(a).$$

3

WolframAlpha distinguishes between x^(1/3) and cbrt(x) - the first returns the principal root while the second returns the real root. For example, cbrt(-8) returns $-2$ while (-8)^(1/3) returns $1 + 1.73205i$. Thus, you can use cbrt in your WolframAlpha query. integrate (1/(3*sqrt(x)*cbrt(log(x))), x=0..2) Note that the output states that the ...

3

Not an answer: (couldn't fit it in comment box) The function $\displaystyle \operatorname{Ti_2}(w) = \int_0^w \frac{\tan^{-1} t}{t}\,dt$ satisfies \displaystyle \operatorname{Ti_2}\left(\frac{w^2}{2}\right) + ... 3 We have: $$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$ while on the other hand, if we set: $$A(n,m)\triangleq\int_{0}^{1} x^n ... 3 You already took care of the endpoints when you did the transformation. No further substitution is needed.$$ \int_1^9 \dfrac{f(\sqrt{x})}{\sqrt{x}}\; dx = 2 \int_1^3 f(u)\; du = 2 \left(\int_1^2 f(u)\; du + \int_2^3 f(u)\; du\right) = 6$$2 We have$$\int \tan(x)dx = -\ln\left|\cos(x)\right|+C$$hence$$\int_{-\pi/3}^{\pi/3} \tan(x)dx = -\ln\left|\cos\left(\frac{\pi}{3}\right)\right|+\ln\left|\cos\left(\frac{-\pi}{3}\right)\right| \\ = -\ln\left|\frac{1}{2}\right|+\ln\left|\frac{1}{2}\right| \\ = 0$$so you are right. Looks like the book probably incorrectly evaluated to ... 2 Hint. Recall that the incomplete gamma function may be defined as$$ \Gamma(s,a)=\int_a^{\infty} t^{s-1}e^{-t}dt,\quad a>0, s \in \mathbb{R}. \tag1 $$Observe that, for 0<x<1, we have$$ \begin{align} \int_0^\infty dn \,\frac{x^n}{3n+1}&=\int_0^\infty (3n+1)^{-1}x^n \:dn\\\\ &=x^{-\frac13}\int_{0}^\infty (3n+1)^{-1}x^{\frac13 (3n+1)} ... 2 \begin{align} \int_{0.5\pi}^{\pi}\frac{1}{1-\cos x}dx &=& \int_{0.5\pi}^{\pi}\frac{1}{2\sin^2\left(\frac{x}{2}\right)}dx\\ \end{align} Here I used $$\cos(2x) = 1-2\sin^2x\implies \cos(x) = 1 -2\sin^2\left(\frac{x}{2}\right)$$ using the subx= 2u$$\int \csc^2u du = -\cot u + C$$ thus \begin{align} -\cot u + C &=& ... 2 Make the two substitutions u=t^2 and y =u -n\pi to findI =\int_{n\pi}^{(n+1)\pi} \frac{\sin(u)}{2\sqrt{u}}du = \int_{0}^{\pi} \frac{\sin(y)(-1)^n}{2\sqrt{y+n\pi}}dy$$where we have used \sin(y+n\pi) = \sin(y)(-1)^n. Now, for y\in[0,\pi] we have that$$\frac{1}{\sqrt{(n+1)\pi}}\leq \frac{1}{\sqrt{n\pi +y}} \leq \frac{1}{\sqrt{n\pi}}$$and also ... 2 The problem is that the logarithm has a branch point in 0, as well as \sqrt[3]{z}, so we must be careful when defining the branches we are taking. If we simply set x=e^t we have:$$ I = \int_{-\infty}^{\log 2}\frac{e^{t/2}}{3 t^{1/3}}\,dt = \int_{0}^{\log 2}3t^{-1/3}e^{t/2}\,dt+\int_{-\infty}^{0}3t^{-1/3}e^{t/2}\,dt\$ where the first integral can be ...

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