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2

Hint: (This expands the hint of @Bernard... it was what I needed to make it work.) Note that $\frac{1}{1-\tan^{2}(\theta)}=\frac{1+\tan^{2}(\theta)}{(1-\tan^{2}(\theta))(1+\tan^{2}(\theta))}=\frac{\sec^{2}(\theta)}{1-\tan^{4}(\theta)}$. So using the suggested substitution $u=\tan(\theta)$ gives you $$\int \frac{\tan^{4}\theta}{1-\tan^{2}\theta}d\theta= ...


1

Suppose the integral were $I=\int_0^{\infty} e^{-y^2-\frac{x^2}{y^2}}dy$. Then we note that $y^2+\frac{x^2}{y^2}=\left(y-\frac{|x|}{y}\right)^2+2|x|$. Thus, we have $$I=e^{-2|x|}\int_0^{\infty}e^{-\left(y-\frac{|x|}{y}\right)^2}dy \tag 1$$ Now, substitute $y\to |x|/y$ so that $dy\to -|x|dy/y^2$. Then, ...


0

Well, several solutions are already given. I will add another one. It is based on a technique called 'trigonometric substitution'. Let $x=\sec \theta ,$ then $$dx=\sec \theta \tan \theta d\theta $$ and $$\sqrt{x^{2}-1}=\sqrt{\sec ^{2}\theta -1}=\sqrt{% \tan ^{2}\theta }=\tan \theta .$$ [ Here $\theta $ is such that $\tan \theta >0$]. Note that $x=\sec ...


4

Hint: Use the substitution $u=\tan\theta$, $ u=(1+u^2)\mathrm d\mkern1mu\theta$. You'll get the integral of the rational function: $$\int\frac{u^4}{1-u^4}\,=\int\frac{u^4}{(1-u)(1+u)(1+u^2)}\,\mathrm d\mkern1mu u$$ Then, decomposition in partial fractions and back to $\theta$.


0

HINT: Multiply the numerator and denominator by $\cos^2(\theta)$. Rewrite $\cos^2(\theta) - \sin^2(\theta)$ as $\cos(2\theta)$ and $\sin^2(\theta)$ in terms of $cos(2\theta)$. We know how to integrate $\sec(2\theta)$.


0

We can find , when $\displaystyle\frac{d}{dx}\cos^{-1}\frac 1x=\frac{d}{dx}\sec^{-1}x$, as shown here: $$y=\sec^{-1}x$$ $$\sec y=x$$ $$\frac{d(\sec y)}{dx}=1$$ Now from differential calculus we should know that $(\sec x )'=\sec x\tan x$, so $$\sec y \tan y \cdot y'=1$$ $$y'=\frac{1}{\sec y \tan y}$$ $$y'=\frac{1}{\sec y\sqrt{\sec^2 y-1}}$$ ...


0

If you are going to use Parseval, then you need to evaluate $a_{n}=\frac{1}{\pi}\int_{-\pi }^{\pi }\sin ^{2}t\cos ntdt$. Since the integrand is even, it suffices to integrate from $0$ to $\pi $ and multiply by $2$. $b_{n}$ is a similar integral with $\sin nt$.But the integrand in this case is odd, so these terms vanish. So you are left with the integrals ...


0

HINT: set $$t=\sqrt{x^2-1}$$ then you will get $$dx=\frac{t}{x}dt$$


3

Notice that $$\int{\frac{dx}{x\sqrt{x^{2}-1}}}= \int{\frac{1}{\sqrt{1-(\frac{1}{x})^{2}}}\frac{dx}{x^{2}}}= \int{\frac{-d(\frac{1}{x})}{\sqrt{1-(\frac{1}{x})^{2}}}} = -\arcsin(\frac{1}{x}) + C$$ or if you prefer you can make the change of variable $y=\frac{1}{x}$


0

Well, differentiate: $$\left(\arccos\frac1x\right)'=\frac1{x^2}\frac1{\sqrt{1-\frac1{x^2}}}=\frac1{x\sqrt{x^2-1}}$$ so what your homework says is correct.


0

I would think that if $x>1$, $$\frac{d}{dx} \sec^{-1}{x} = \frac{1}{x\sqrt{x^2-1}}$$ so $$\int \frac{1}{x\sqrt{x^2-1}} = \sec^{-1}{x}, x>1$$ Edit: Since you have so many answers, note that to say the antiderivative is $\cos^{-1}{(1/x)}$ and that it is $\sec^{-1}{x}$ is equivalent because cosine and secant are reciprocal functions. If you have the ...


5

We have: $$\sin^2(x) = \frac{1-\cos(2x)}{2}\tag{1}$$ hence Parseval's identity implies: $$ \int_{-\pi}^{\pi}\sin^4(x)\,dx = 2\pi\cdot\frac{1}{4}+\pi\cdot\frac{1}{4}=\color{red}{\frac{3\pi}{4}}.\tag{2}$$


0

HINT: $\sin(x)^4=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$


3

I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$\int_0^\infty \exp^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other ...


0

Here is the answer done by Maple: int(sqrt(tan(x)), x); $$\frac 1 2\,{\frac {\sqrt {2}\sqrt {\tan \left( x \right) }\cos \left( x \right) \arccos \left( \cos \left( x \right) -\sin \left( x \right) \right) }{\sqrt {\cos \left( x \right) \sin \left( x \right) }}}-$$ $$\frac 1 2 \,\sqrt {2}\ln \left( \cos \left( x \right) +\sqrt {2}\sqrt {\tan \left( x ...


0

As already mentioned in some answers, let $t^2=\tan x \implies 2tdt=\sec^2x dx$ or $dx=\frac{2t}{t^4+1}$Now, We can easily reach to the final answer as follows $$I=\int \frac{2t^2 dt}{t^4+1}=\int \frac{2 dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}$$ $$=\int ...


2

Since: $$\frac{d}{dx}\sqrt{1+x^4} = \frac{4x^3}{2\sqrt{1+x^4}}$$ we have: $$ \int_{0}^{1}\frac{x^3}{\sqrt{1+x^4}}\,dx = \frac{\sqrt{2}-1}{2}.$$


1

hint What about the change of variable $1+x^4=u$ such that $4x^3dx=du$ So the integral becomes $$\frac{1}{4}\int_1^2 u^{-\frac{1}{2}}du$$ Can you take it from there


0

Notice that: $$\frac{\sin^3(x)}{2+\cos x} = -\left(\frac{d}{dx}\cos x\right)\frac{1-\cos^2 x}{2+\cos x}$$ hence you just need to integrate: $$\frac{1-z^2}{2+z} = 2-z-\frac{5}{2+z},$$ not a difficult task.


2

HINT: Write $\sin^3x=\sin x(1-\cos^2x)$ and set $2+\cos x=y\iff\cos x=\cdots$


4

The General Case: Consider instead the integral \begin{align} J(\eta) &=\Re\int^{2\pi}_0x^\eta\ {\rm Li}_2(e^{ix})^2\ {\rm d}x\\ \end{align} for $\eta\in\mathbb{N}_0$. Deforming the contour around $z=0$, we get \begin{align} J(\eta) &=\Re\oint_{|z|=1}\frac{{\rm Li}_2(z)^2\ln^{\eta}(z)}{i^{\eta+1}z}{\rm d}z\\ &=\Re\int^1_0\frac{\left((\ln(x)+2\pi ...


3

$I(1)$ can be evaluated in the following way also. One has to note that $$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\begin{cases} 0 & n \neq m \\ \pi^2 & n=m \end{cases}$$ To prove the above, write $$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\frac{1}{2}\int_0^{2\pi} x\cos((m-n)x) \,dx-\frac{1}{2}\int_0^{2\pi} x\cos((m+n)x)\,dx$$ and then use integration by ...


1

This antiderivative is not an elementary function. However, at $\theta = \pi$ the integral is $\pi I_0(a)$ where $I_0$ is a modified Bessel function. EDIT: With the substitution $x = \arcsin(t)$, the integral becomes $$ I = \int_0^{\sin(\theta)} \dfrac{\cosh(at)\; dt}{\sqrt{1-t^2}} = \sum_{k=0}^\infty {2k \choose k} 4^{-k} \int_0^{\sin(\theta)} t^{2k} ...


0

Note that $x^4+1=(x^2+i)(x^2-i)$. Then, using partial fractions, \begin{align} \int\frac{1}{x^4+1}dx=\frac{1}{2}i\int\frac{1}{x^2+i}dx-\frac{1}{2}i\int\frac{1}{x^2-i}dx \end{align} Use partial fractions again, then you should be able to integrate to find something involving $\ln(\cdot)$ plus other terms. Alternatively, instead of using partial fractions a ...


0

Hint: we have $$x^4+1=-\left(-x^2+\sqrt{2} x-1\right) \left(x^2+\sqrt{2} x+1\right)$$


1

Numerically, I get $$ S_1+S_2 = 0.14836252987273216621 $$ which agrees with $$ \frac{\pi^6}{6480} $$ Also numerically, $$ S_1 = 0.074181264936366083104 \\ S_2 = 0.074181264936366083104 $$ are seemingly equal.


2

No. The function $$ f(x)=\begin{cases}0 & |x|>1\\ (1-\log(1-x^2))^{-1} & |x|\le1\end{cases} $$ is an explicit example. For more comments and a proof (due to Terry Tao) see this post.


2

No, there exist continuous functions on $S^1$ whose Fourier coefficients are not in $l^{\beta}$ for every $\beta <2$ ( while certainly being in $l^2$). The first example seems to be given by T. Carleman in 1916. This should be discusses in some books on classical Fourier series. I learned about it from this paper of Hausdorff.


3

Hint You arrived to $$I=\int\frac{\cos^2(\theta)}{1+\sin^2(\theta)}d\theta$$ Use the double angle formula to express $\cos^2(\theta)$ and $\sin^2(\theta)$ as function of $\cos (2\theta)$. So, $$I=\int\frac{1+\cos (2\theta)}{3-\cos (2\theta)}d\theta$$ Now, use the tangent half-angle substitution $t=\tan(\theta)$ $$I=\int\frac{2}{2 t^4+3 t^2+1}dt$$ The roots ...


0

The disk method can be used here. $V = \displaystyle \int_{-1}^1 \pi\left((2-x^2)^2-1^2\right)dx$


0

at $x=5$ $$20.1=4.5+a(5)^3$$ $$a=0.1248$$ now we will find the Area $$\text{Area}=\int_{0}^{5}(4.5+0.1248x^3)dx=42$$ now we will find the moment of area about the x=0 $$\text{Moment}=\int_{0}^{5}(4.5+0.1248x^3)(x)dx=134.25$$ then find the center $$C=\frac{\text{Moment}}{\text{Area}}=\frac{134.25}{42}=3.196....$$


4

$$\int\frac{\cos^2\theta}{1+\sin^2\theta}d\theta$$ multiply num and denum by $\sec^4\theta$ $$\int\frac{\sec^2\theta}{\sec^4\theta+\sec^2\theta\tan^2\theta}d\theta$$ trigonometric identity $\sec^2\theta=1+\tan^2\theta$ $$\int\frac{\sec^2\theta}{2\tan^4\theta+3\tan^2\theta+1}d\theta$$ substitute $u=\tan\theta$ The rest is partial fraction expansion + ...


1

when you have so many fractional powers, to simplify things try to go for a substitution that can clear up all the powers (usually the LCM of all denominators of various powers will do the job). For example, here let $x=t^{12}$, then you get \begin{align*} \int {\frac{1}{\sqrt[3]{x} +\sqrt[4]{x}}} +\frac{\log{(1+\sqrt[6]{x})}}{\sqrt[3]{x}+\sqrt{x}}\, dx ...


0

$dv$ is just a mnemonic for $v'dx$ in your first equation, in the second one you are using $v$ in place of $v'$ so $v$ turns into the primitive of $v$, $\int vdx$.


4

Let $t=\frac{1}{x}$, so then $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$. Then $\displaystyle\int\frac{1}{x^{11}+4x^6}dx=\int\frac{1}{\frac{1}{t^{11}}+\frac{4}{t^6}}\left(-\frac{1}{t^2}\right)dt=-\int\frac{t^9}{1+4t^5}dt$ ...


1

Let $R$ be so large that $\Omega\subset B(x,R)$ (where $B(x,R)$ is the ball centered at $x$ and with radius $R$). Then, since the function is positive, you have, by domain-monotonicity, $$ \int_\Omega|x-y|^{1-n}\lambda^n(y)\leq \int_{B(x,R)}|x-y|^{1-n}\lambda^n(y). $$ Now, let $u=x-y$, and you get that the integral to the right equals $$ ...


0

When I'm treating an integral of the form $\int f(x)\>g(x)\>dx$, indefinite or definite, using partial integration I use the following arrow notation: I put a downward arrow $\downarrow$ under the factor that gets differentiated in the process and an upward arrow $\uparrow$ under the factor that gets integrated. This saves me from introducing new ...


2

Multiple choice: $y$ is a constant, i.e. something that doesn't change as $x$ changes, and you're antidifferentiating a function of $x$. Then this becomes $$ -\int\frac{du}{\sqrt{1+u^2}}. $$ It's not altogether impossible that $x$ is a constant and you're antidifferentiating a function of $y$, so that $\displaystyle d\left(\frac x y \right) = ...


1

The notation is a bit unclear to me. But let's say that you want to find the following integral. $$ \int -\frac{1}{\sqrt{1 + (x/y)^2}}\; dx. $$ Here, from the notation, we assume that $y$ is a constant. The standard approach is integration by substitution. So we let $u = x/y$ Then $du = (1/y) dy$. So we get the integral $$ \int -\frac{1}{\sqrt{1 + (u)^2}}\; ...


0

I think you're getting tripped up because the $v$ used in each identity is different. To see why, use each to evaluate $\int x \cos x\,dx$. In the second identity, $v=\cos x$. But in the first identity, $dv=\cos x\,dx$, so $v=\sin x$. Rather, you can prove the second identity by showing that the derivative of the of right-hand side is $uv$. Since ...


1

The $\oint$ is called line integral The terms path integral, curve integral, and curvilinear integral are also used; contour integral as well, although that is typically reserved for line integrals in the complex plane. using it we take perimeter instead of area in contrast to $\int$. This finds great use in physics (ampere circuital law) The $\iint$ ...


2

The first is Contour Integral The second and third are double and triple integrals


1

The first denotes a closed integral. Its meaning varies a little based on context, but the best way to think of it is that the set of points over which you are integrating is closed and bounded (like integrating over the circumference of a circle or the outside of a sphere). There might be a little bit of contention with the leftmost one since physicists ...


5

I can confirm $I(1)$: Using the series expansion for Clausen's function we have $$ I(p)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\int_0^{2 \pi}x^p\frac{\sin(n x)}{n^2}\frac{\sin(m x)}{m^2}\,\mathrm dx $$ now setting $p=1$. Furthermore we may use the orthogonality of sine $\int^{2 \pi}_{0}{\sin(n x)\sin(m x)}\,\mathrm dx=\pi\delta_{mn}$ and integrate by ...


4

$$\int\frac{1}{x^{11}+4x^6}dx=$$ $$\int \left(\frac{1}{4x^6}+\frac{x^4}{16(x^5+4)}-\frac{1}{16x}\right)dx=$$ $$\frac{1}{16}\int \frac{x^4}{x^5+4}dx-\frac{1}{16}\int \frac{1}{x}dx+\frac{1}{4}\int \frac{1}{x^6}dx=$$ $$\frac{1}{80}\int \frac{1}{u}du-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx=$$ ...


4

After using partial fractions, try the substitution $u=x^5+4$ to deal with the first integral in the expression \begin{equation*} \frac{1}{16}\int\frac{x^4}{x^5+4}dx-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx \end{equation*} which is what you get when you use partial fractions on your integrand.


1

Well, the tangent line at $(1,3)$ is parallel to the $x$-axis, meaning that the gradient when $x=3$ is...what?


5

Hint: $$ \begin{align} \int\frac{dx}{x^{11}+4x^6}&=\int\frac{x^4dx}{x^{15}+4x^{10}}\\ &=\frac14\int\frac{du}{u^3+4u^2}\,\,\,\,\,,u=x^5\\ &=\frac{1}{16}\int\bigg[\frac{1}{u^2}-\frac{1}{4u}+\frac{1}{4(4+u)}\bigg]du \end{align} $$


1

$$I=\int_{0}^{a}J_0(b\sqrt{a^2-x^2})\cosh(cx)\,dx = a\int_{0}^{1}J_0(ab\sqrt{1-z^2})\cosh(acz)\,dx$$ The trick is now to expand both $J_0$ and $\cosh$ as Taylor series, then to exploit: $$ \int_{0}^{1}(1-z^2)^{n}z^{2m}\,dz = \frac{\Gamma(n+1)\,\Gamma\left(m+\frac{1}{2}\right)}{2\cdot\Gamma\left(m+n+\frac{3}{2}\right)}\tag{2}$$ hence, by assuming $b>c$ and ...



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