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1

So, you know the two curves intersect at $p = \cos^{-1} (1/2)$, which implies $p = \pm \pi/3 + 2\pi k$ for any $k \in \mathbb{Z}$. For which $k$ do we observe an intersection within the specified integral of integration? That is, for which $k$ is $p \in [0, \pi/7]$? (Hint: there is only one such $k$) The last step is to integrate $f(x) - g(x)$ from $0$ to ...


0

As was mentioned: the integrand simplifies to: $\sqrt{a^2 + b^2 - 2 a b \cos\theta \sin\phi}$ Factor out $\sqrt{a^2 +b^2}$ and use the binomial theorem to expand, treating $\cos\theta \sin \phi$ as small. This gives: $$\sqrt{a^2 +b^2} \left(1-\frac{a b \cos (\theta ) \sin (\phi )}{a^2+b^2}-\frac{a^2 b^2 \cos ^2(\theta ) \sin ^2(\phi )}{2 ...


3

We will use similar approach as sos440's answer in I&S. Using the simple algebraic identity $$ ab^2=\frac{(a+b)^3+(a-b)^3-2a^3}{6}, $$ it follows that \begin{align} \int_0^1 \frac{\ln x\ln(1+x)\ln^2(1-x)}{x}\ dx &=\frac16I_1+\frac16I_2-\frac13I_3\ ,\tag1 \end{align} where \begin{align} I_1&=\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}\ dx\\[12pt] ...


2

$$I=\int_{-2\pi}^{2\pi} xe^{-\mid x\mid}dx.$$ Loosely speaking, you can think about the definite integral as the area bounded by the function $xe^{-\mid x\mid}$ and the $x$-axis, as the variable $x$ moves from $x=-2\pi$ to $x=0$ then from $x=0$ through to $x=2\pi$. So, intuitively it's not too much of a step to see that $$I=\int_{-2\pi}^{0} xe^{-\mid ...


1

If you have a function $f: [a,b] \to \Bbb R$ defined by parts, as in: $$f(x) = \begin{cases} f_1(x), \mbox{if } a \leq x \leq c \\ f_2(x), \mbox{if } c < x \leq b\end{cases}$$ then: $$\int_a^b f(x) \ \mathrm{d}x = \int_a^c f_1(x) \ \mathrm{d}x + \int_c^bf_2(x) \ \mathrm{d}x.$$ In your case, $f(x) = xe^{-|x|}$, so $c = 0$ and $f_1(x) = xe^{x}$ and $f_2(x) ...


2

$$\int_{-2\pi}^{2\pi}xe^{-\left|x\right|}dx=\int_{-2\pi}^{0}xe^{-\left|x\right|}dx+\int_{0}^{2\pi}xe^{-\left|x\right|}dx=\int_{-2\pi}^{0}xe^{x}dx+\int_{0}^{2\pi}xe^{-x}dx$$ This is a split of cases killing the annoying modulus.


0

$|x|=x$ for x>0 and = -x for x<0 and 0 for x=0 since x is positive and negative in the limits $-2\pi$ to $2\pi$ you have to split it into two parts and then evaluate your integral. So in this function you are splitting into cases for |x| where x>0 and x<0. this is done because |x| is defined that way in its domain.since you get two different ...


1

I do not think taht any closed form could be found for these integrals except in the case where $b^2-4ac=0$ where we should find $$a x^2+bx+x=a \Big(x+\frac{b}{2a}\Big)^2$$ $$\sqrt{a x^2+bx+x}=\sqrt a (x+\frac{b}{2a})$$ For such a case $$\displaystyle\int\frac{\sin^2\left( \sqrt{ ax^2+bx+c}\right) }{ \sqrt{ ax^2+bx+c}} dx =\frac{\log \left(\frac{2 a x+b}{2 ...


7

Here are values of the integral $[3]$ for some specific values of the parameter $a$: ...


1

Let $u=\sin^{-1}t,\; dv=t dt$, so $\;du=\frac{1}{\sqrt{1-t^2}}dt, \;v=\frac{t^2}{2}$. Then $\displaystyle\int t\sin^{-1}t \;dt=\frac{t^2}{2}\sin^{-1}t-\frac{1}{2}\int\frac{t^2}{\sqrt{1-t^2}} dt$. Now let $t=\sin\theta, dt=\cos\theta d\theta$ to get $\displaystyle\int\frac{t^2}{\sqrt{1-t^2}} dt=\int\sin^{2}\theta d\theta$, and then use a half-angle ...


0

As you say, there is a discontinuity, so let us just consider $\int_4^{16} \frac {\sqrt x}{x-4}dx$ to start. If we substitute $u=x-4$ this becomes $\int_0^{12}\frac{\sqrt {u+4}}{u}du$ Very near $u=0$ this is essentially $\int_0^\epsilon \frac 2u du$ which diverges logarthmically, so you can just report that the integral does not exist. There is similar ...


1

Proper substitution: Draw the right triangle with one side being $t$ and a hypotenuse of $1$, the other side is then $\sqrt{1-t^{2}}$. Now substitute: $$\theta =Sin^{-1}(t)\Rightarrow \frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{1}{\sqrt{1-t^{2}}}\Rightarrow dt=\sqrt{1-t^{2}}d\theta =Cos(\theta )d\theta $$ $$\int tSin^{-1}(t)dt=\int \theta Sin(\theta ...


0

Another (probably) easier solution $$ 1+x^2+16x^4 \geq 1+x^2 \implies \frac{1}{1+x^2+16x^4} \leq \frac{1}{1+x^2} \implies \\ \int_{1}^{+\infty}\frac{dx}{1+x^2+16x^4} \leq \int_{1}^{+\infty}\frac{dx}{1+x^2} = \frac{\pi}{4} $$


0

By AM-GM $1+16x^4\geq 8x^2$, hence: $$\int_{1}^{+\infty}\frac{dx}{1+4x^2+16x^4}\leq\int_{1}^{+\infty}\frac{dx}{12x^2}.$$


1

Just see this if $x\geq 1$ $$ x^2 \geq x \implies x^4 \geq x^2 \implies 16 x^4 \geq 4x^2 \implies 1+x^2+16 x^4 \geq 4x^2 $$ since you adding positive terms. So can you see it now?


0

Hint: Clearly, $x>6>1>0$, therefore $\underbrace{\sqrt{0+x^2}}_x<\sqrt{1+x^2}<\underbrace{\sqrt{x^2+x^2}}_{x\sqrt2}$.


8

First, we set $x=\cos\theta\ \color{red}{\Rightarrow}\ dx=-\sin\theta\ d\theta$, then \begin{align} \mathcal{I}=\int_0^1\frac{\ln x}{1+x}\arccos(x)\ dx=-\int_0^{\Large\frac{\pi}2}\frac{\sin\theta}{1+\cos\theta}\cdot\theta\ \ln (\cos\theta)\ d\theta.\tag1 \end{align} Using the fact that \begin{align} ...


1

To solve such problems, always plot the given curves. This will help you figure out what integrals you need to evaluate. For any $x=x_0$ the area will be defined by two enclosing curves; one above and one below. First note that the are defined by the three curves begins wherever $x^2+1=-4x-3$. Let's call that point $x_a$. For $x_a \le x \le 0$ The first ...


3

Hint: Here is a diagram of the situation. Calculate the area of the triangle, and then subtract the area formed by the parabola. The area will be $\displaystyle A=16-\int_{-2}^25-(1+x^2)\,dx$. I think you can take it from here. If not, let me know. Edit First, we take the area of the whole triangle, which is $4\cdot8\cdot\frac12=16$ Then, we ...


2

If you evaluate the integral using the substitution $$x=tan(\theta)$$ the result is integration of $$sec(\theta)$$ with changed limits of integration and if evalueted completely it tends to infinity


6

You have, for $x>0$, $$ \sqrt{1+x^2}\leq \sqrt{1+2x+x^2}= \sqrt{(x+1)^2}=x+1 $$ thus $$ \int_6^\infty\frac{1}{\sqrt{1+x^2}}dx\geq \int_6^\infty\frac{1}{x+1}dx=\infty $$ and your integral is divergent.


4

Notice that $$\frac{1}{\sqrt{1+x^2}}\sim_\infty\frac 1x$$ so the given integral is divergent.


2

Hint: By the change of variable $$P=x^8, \qquad dP=8x^7dx$$ you readily obtain $$ \begin{align} \nu &=\int_{0}^{P(r)} \,\frac{dP}{P+\beta\rho(P)}\\\\ &=8 \int_{0}^{\sqrt[8]{P(r)}} \,\frac{x^7}{x^8+\beta \left(0.0019x^6+0.002x^5+0.003x^4\right)}{\rm d}x\\\\ &=8 \int_{0}^{\sqrt[8]{P(r)}} \,\frac{x^3}{x^4+0.0019x^2+0.002x+0.003}{\rm d}x\\ ...


1

Since the function $e^{-4|x|}$ is even we have $$\int_{-\infty}^{\infty}e^{-4|x|}\, dx=2\int_{0}^{\infty}e^{-4x}\, dx$$ On the other hand $$2\int_{0}^{\infty}e^{-4x}\, dx=2\lim\limits_ {a\to \infty }\int_{0}^{a}e^{-4x}\, dx=\frac{1}{2}$$


4

$$\int_{-\infty}^{\infty}e^{-|4x|}dx=\int_{-\infty}^{0}e^{-|4x|}dx+\int_{0}^{\infty}e^{-|4x|}dx$$ $$=\int_{-\infty}^{0}e^{4x}dx+\int_{0}^{\infty}e^{-4x}dx=2\int_{0}^{\infty}e^{-4x}dx=1/2$$


0

Your derivative is wrong. While it would be a worthwhile exercise to work carefully how you were computing the derivative and figure out where you went wrong, an easier approach for just getting the answer is to realize you can simplify away all of the appearances of the absolute value.


1

First look at the antiderivative $$I=\int\sin^2(x)\cos^4(x)\hspace{1mm}dx$$ and now use $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$ so $$\cos^4(x)=\Big(\frac{\cos(2x)+1}{2}\Big)^2=\frac{1}{4}\Big(\cos^2(2x)+2 \cos(2x)+1\Big)$$ $$\cos^4(x)=\frac{1}{4}\Big(\frac{\cos(4x)+1}{2}+2 \cos(2x)+1\Big)$$ ...


1

You got the right answer. Not all integrals are easy to solve. Some are tricky and take time. If you want to see some examples of this, get the new book by Paul J. Nahin "Inside Interesting Integrals".


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


2

For $(3)$ you can have the form $$ I = - \rm Li_3( -2 ) \ln\left( {\frac {a+2}{2}} \right) - {\rm Li_4}( a )+ \int _{0}^{a}\,{\frac {{\rm Li_3}(t) }{t+2}}{dt},$$ where $\rm Li_s(z)$ is the polylogarithm function. For instance when $a=1/2$ we have $$ I \sim - 0.08900930960. $$


0

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5

Letting $ \displaystyle f(z) = \frac{\ln^{2}z}{\sqrt{z} (z-1)} $ and integrating around a keyhole contour where the branch cut is along the positive real axis, $$\int_{0}^{\infty} \frac{\ln^{2} x}{\sqrt{x}(x-1)} \ dx + \text{PV} \int_{\infty}^{0} \frac{(\ln x + 2 \pi i )^{2}}{\sqrt{x e^{2 \pi i}}(x-1)} \ dx - i \pi \ \text{Res} [f(z), e^{2 \pi i}] =0 $$ ...


1

Taking into account the fact that you already received very good answers, I give here just as a curiosity (given by a CAS), $$\int \frac{\ln(x)}{\sqrt{x}(x-1)} dx=-2 \text{Li}_2\left(1-\sqrt{x}\right)-2 \text{Li}_2\left(-\sqrt{x}\right)-\log \left(\sqrt{x}+1\right) \log (x)$$ $$\int_0^a \frac{\ln(x)}{\sqrt{x}(x-1)} dx=\sqrt{a} \left(a \Phi ...


4

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0

It depends on your definition/understanding of "definite integral". If you think of the Lebesgue definite integral, the answer is yes, because each function get mapped to a scalar value and this action is linear. However, this action may not be bounded. There are, however, more general ideas of definite integrals, e.g., the Bochner integral. Think of a ...


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


4

Make the substitution $\sqrt{x}=t$ which gives $$4\int_{0}^{\infty}{\frac{\log(t)}{t^{2}-1}dt}$$ Notice the symmetry at $t=1$, or if you prefer spliting up the integral and making the change of Variable $t=\frac{1}{u}$ at $(1,\infty]$ This reduces to $$8\int_{0}^{1}{\frac{\log(u)}{u^{2}-1}du}$$ Expanding the denomenator into a geometric series and applying ...


6

Another (low-tech) way to prove your claim is to consider that: $$\begin{eqnarray*} I &=& 4\int_{0}^{+\infty}\frac{\log x}{x^2-1}\,dx =4\left(\int_{0}^{1}\frac{\log x}{x^2-1}\,dx + \int_{1}^{+\infty}\frac{\log x}{x^2-1}\,dx \right)\\&=&8\int_{0}^{1}\frac{-\log x}{1-x^2}\,dx=8\sum_{k=0}^{+\infty}\int_{0}^{1}(-\log ...


6

Related problems: (I). We will use the Mellin transform approach $$ F(s) = \int_{0}^{\infty} x^{s-1}f(x)dx \implies F'(s)= \int_{0}^{\infty} x^{s-1}\ln(x)f(x)dx. $$ So the whole problem boils down to find the Mellin transform of $\frac{1}{x-1} $ which is given by $$ F(s) = (-1)^{s+1}\Gamma(s)\Gamma(1-s) , $$ where $\Gamma(z)$ is the gamma ...


7

$$\int_0^1\frac{\ln x\cdot\arccos x}{1+x}\,dx=2\,G\ln2+\frac\pi2\ln^22+\frac{\pi^3}{16}-4\,\Im\operatorname{Li}_3(1+i).$$ You also can have a different form that does not use complex numbers: $$\int_0^1\frac{\ln x\cdot\arccos ...


0

This is what $\sinh^{-1}x$ has to do with $\tan^{-1} x$. $$\frac{1}{x^2+1} = \bigg(\frac{1}{\sqrt{x^2+1}}\bigg)^2$$ $$\frac{d}{dx}\tan^{-1}x = \bigg(\frac{d}{dx}\sinh^{-1}x\bigg)^2$$ This fact (that both derivatives are based on $x^2+1$) comes from the similarity in the following identities $$\sec^2 x=\tan^2x +1$$ $$\cosh^2 x = \sinh^2 x + 1$$


1

It's actually enough to resolve the $n=0$ case, since $t$-derivatives of $$F(t)=\int_0^\infty \frac{e^{i x t}}{\sqrt{e^{2x}-1}} x^{1/2}\,dx$$ will bring down more powers of $x$; I've chosen units for so that $a=1$ for convenience. Even with these simplifications, I don't know how to compute a closed-form and so will have to settle for an appropriate series ...


0

$x^{n+\frac{1}{2}} (1-e^{-a x })^{-\frac{1}{2}} e^{x(-\frac{a}{2})} dx=x^{n+\frac{1}{2}}(\sum_p\frac{(2p-1)!!}{2p!!}e^{-a(p+\frac{1}{2})x})$. So $\int x^{n+\frac{1}{2}} (1-e^{-a x })^{-\frac{1}{2}} e^{x(-\frac{a}{2})} dx=\Gamma(n+1.5).\sum_p \frac{(2p-1)!!}{2p!!}\frac{1}{a(p+\frac{1}{2})}^{n+1.5}$ But I guess you already had this, and this is not really ...


3

A bit convoluted, but anyway... Assume $0\leq x\lt1$, and write $$f(x)=\sqrt{1+\sqrt{1-x^2}}$$ $$g(x)=\sqrt{1-\sqrt{1-x^2}}$$ First, notice that $$f(x)g(x)=\sqrt{1-1+x^2}=x$$ Let's differentiate $f^2(x)=1+\sqrt{1-x^2}$: $$2f'(x)f(x)=-\frac{x}{\sqrt{1-x^2}}$$ So $$f'(x)=-\frac{x}{2f(x)\sqrt{1-x^2}}$$ Likewise, ...


3

Substitute $x=\sin(2u)$: $$ \begin{align} \int_0^{1/2}\sqrt{1+\sqrt{1-x^2}}\,\mathrm{d}x &=\int_0^{\pi/12}\sqrt{1+\cos(2u)}\,\mathrm{d}\sin(2u)\\ &=2\int_0^{\pi/12}\sqrt{2\cos^2(u)}\cos(2u)\,\mathrm{d}u\\ &=\sqrt2\int_0^{\pi/12}(\cos(u)+\cos(3u))\,\mathrm{d}u\\ &=\sqrt2\left[\sin(u)+\frac13\sin(3u)\right]_0^{\pi/12}\\ \end{align} $$ Then use ...


6

Split the integral into two forms by expanding the logarithm function $$ \int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^1\frac{\ln(3+x)}{\sqrt{x(1-x)}}\ dx-\int_{0}^1\frac{\ln(3-x)}{\sqrt{x(1-x)}}\ dx $$ Let $t=\sqrt{x}\ \rightarrow\ dt=\dfrac{dx}{2\sqrt{x}}$, we have $$ 2\int_{0}^1\frac{\ln(3+t^2)}{\sqrt{1-t^2}}\ ...


6

$$x=\sin2y\implies1-x^2=\cos^22y\implies1+\sqrt{1-x^2}=1+\cos2y=2\cos^2y$$ $$\sin2y=\dfrac12,y=\frac\pi{12};\sin2y=0,y=0$$ $$\int_0^1\sqrt{1+\sqrt{1-x^2}}dx=\sqrt2\int_0^\dfrac\pi{12}\cos y(2\cos2y)dy$$ $$\sqrt2\int_0^\dfrac\pi{12}\cos y(2\cos2y)dy=2\sqrt2\int_0^\dfrac\pi{12}\cos y(1-2\sin^2y)dy$$ Set $\sin y=u$


3

$$ \int \sqrt{1+\sqrt{1-x^2}}dx $$ let $u^2=1-x^2$ $$ \begin{align} \int \sqrt{1+u}\dfrac{u}{-x}du &=& -\int \sqrt{1+u}\dfrac{u}{\sqrt{1-u^2}}du\\ &=&-\int \dfrac{u\sqrt{1+u}}{\sqrt{(1-u)(1+u)}}udu\\ &=& = -\int\dfrac{u}{\sqrt{1-u}}du \end{align} $$ then let $v = 1-u$ to be more clear $$ -\int\dfrac{u}{\sqrt{1-u}}du = -\int ...


10

We have: $$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx = 2\pi\log\frac{2+\sqrt{3}}{2}.\tag{1}$$ This happens because: $$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=2\int_{0}^{1}\frac{\log(3+x^2)}{\sqrt{1-x^2}}\,dx =\int_{-\pi/2}^{\pi/2}\log(3+\cos^2\theta)\,d\theta,$$ ...


12

Let $x = \sin(t)^2$ and $s = 2t$, we have $$\int_0^1 \log\left(\frac{3+x}{3-x}\right)\frac{dx}{\sqrt{x(1-x)}} = \int_0^{\pi/2} \log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right)\frac{2\sin t\cos tdt}{ \sqrt{\sin(t)^2(1-\sin(t)^2)}}\\ = 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right) dt = \int_0^{\pi}\log\left(\frac{3 + \frac{1-\cos ...



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