9

A correct antiderivative is $$ \int\frac{1}{\sin^4 x+\cos^4 x}\, \mathrm{d}x = \frac{1}{\sqrt{2}}\arctan\left(\frac{1}{\sqrt{2}}\tan(2x)\right)+\mathsf{C}, $$ which is true on every interval $ \left(\frac{k\pi}{2}-\frac{\pi}{4}, \frac{k\pi}{2}+\frac{\pi}{4} \right) $ with $k \in \mathbb{Z}$. You may check the following plot for the comparison of both sides....


7

There is no minimum if $x=y=z=t\rightarrow +\infty$ than your expression is $3t^2-2t^3 \rightarrow -\infty$


5

You cannot directly use the indefinite integral to compute the definite integral because $\tan 2\, x$ is not defined at $\pi/4, 3\pi /4, 5\pi /4, 7\pi /4$. But you can use the indefinite integral to compute the definite integral from $0$ to $\pi/4 -\epsilon$ , the integral from $\pi/4+\epsilon$ to $3\pi/4 -\epsilon$ etc. When you take the limit as $\...


5

If $s_n =\sum_{k=0}^n \dfrac1{\sqrt{n+k}} $ then $s_n \ge \sum_{k=0}^n \dfrac1{\sqrt{2n}} =(n+1)\dfrac1{\sqrt{2n}} \gt\dfrac{n}{\sqrt{2n}} =\dfrac{\sqrt{n}}{\sqrt{2}} \to \infty $


4

The volume of water is changing linearly, but the height and volume are related nonlinearly. That is why $h(t)$ is non-linear. \begin{eqnarray*} V &=& \frac{1}{3} \pi r^2 h\\ r &=& h \tan \theta\\ V &=& \frac{1}{3} \pi \tan^2 \theta h^3\\ \frac{dV}{dt} &=& \frac{1}{3} \pi \tan^2 \theta 3 h^2 \frac{dh}{dt}\\ \end{eqnarray*} $\...


4

Define the function $F$ for $x>0$ by: \begin{align}F(x)=\text{cotanh}\left(\frac{x}{2}\right)-\frac{2}{x}\end{align} Observe that, \begin{align}\lim_{x\rightarrow 0} F(x)&=0\\ \lim_{x\rightarrow \infty} F(x)&=1\\ F(x)-F(2x)&=\frac{1}{\sinh x}-\frac{1}{x} \end{align} On can use Frullani's theorem: \begin{align}\int_0^\infty \frac{\text{csch}(...


4

The issue here is what an indefinite integral means. An indefinite integral $\int f(x)dx$ is not a single function; rather, it's best thought of as a set of functions, namely, the set $$\{F: {d\over dx}F=f\}.$$ Elements of this set are called antiderivatives, or primitives, of $f$. This set will always contain more than one element, since adding a constant ...


4

The idea is to use the first part of the FTC. Note that $F=f\circ g$, where $$f(x)=\int_1^x \cot(t)\,dt \qquad \text{and}\qquad g(x)=4x.$$ Then $$f'(x)=\cot(x) \qquad \text{and}\qquad g'(x)=4.$$ Therefore, by the chain rule, the derivative of $F(x)$ is $f'(g(x))\,g'(x)$. That is, $$F'(x)=\cot(4x)\cdot4.$$


4

Hint: For each $k=0,1,\dots,n$, one has $\;\frac 1{\sqrt{n+k}}\ge\frac1{\sqrt{2n}}$. How many such terms do you have? Deduce a lower bound for the sum of these terms.


4

Just an alternate method of evaluation that relies heavily on symmetry. $$J=\int_0^{2\pi}\frac{dx}{\sin(x)^4+\cos(x)^4}$$ We can do a bunch of trig and notice that this simplifies to $$J=4\int_0^{2\pi}\frac{dx}{3+\cos(4x)}=16\int_0^{\pi/2}\frac{dx}{3+\cos(4x)}.$$ Preforming $4x\mapsto x$ gives $$J=4\int_0^{2\pi}\frac{dx}{3+\cos x}=8\int_0^\pi \frac{dx}{3+\...


3

Given integral $$ I = \int^{2 \pi}_0 \cfrac{1}{\sin^4 x + \cos^4 x} dx = 4 \int^{\pi/2}_0 \cfrac{dx}{\sin^4 x + \cos^4 x} $$ By repeatedly using $ \int^{2a}_0 f(x) dx = \int^a_0 f(x) dx + \int^a_0 f(2a-x) dx$ On multiplying numerator and denominator by $ \sec^4(x)$, followed by substitution of $\tan(x) = u$ $$ \begin{align} I &= 4 \int^{\infty}_0 \...


5

I understand that $\frac {dh}{dt}$ is decreasing. However, I don't understand on an intuitive level why $\frac {dh}{dt}$ is non-linear. Because any strictly decreasing function that is linear eventually becomes negative, but you already know that $\frac {dh}{dt}$ is always positive.


3

$$x^4+y^4=c$$ $$4x^3+4y^3 \frac {dy}{dx}=0$$ $$\frac {dy}{dx}= \frac {-x^3}{y^3}$$ For the orthogonal curves we get $$\frac {dy}{dx}=\frac {y^3}{x^3}$$ $$\frac {dy}{y^3}= \frac{dx}{x^3 }$$ $$\frac {-1}{y^2}=\frac {-1}{x^2}+c $$ $$y^2=\frac {x^2}{1+cx^2}$$


3

The gradient of a function $f$ is already perpendicular to all contour curves $f(x, y) = c$. So what you really want is for $r'$ and $\nabla f$ to be parallel, not perpendicular. And yes, a straight line $r(t) = (t, t)$ will do that, as $r'(t) = (1, 1)$ is parallel to $\nabla f(x, y) = (4x^3, 4y^3)$ at any point along $r$ (except perhaps at the origin, that'...


3

The given antiderivative, $$\frac{1}{\sqrt 2} \arctan \left( \frac{1}{\sqrt 2} \tan 2 x \right)$$ is not defined everywhere on the interval $[0, 2 \pi]$ of integration---it is undefined at $\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$, so the usual hypotheses of the Fundamental Theorem of Calculus are not satisfied for that interval. In ...


2

Divide it to two improper integrals with one singularity each. $\int_{0}^{\infty}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx=\int_{0}^{1}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx+\int_{1}^{\infty}\displaystyle\frac{1-cos{x}}{x^\alpha} \,dx$ The original integral will converge $\iff$ Both those integrals will converge $\int_{1}^{\infty}\displaystyle\...


2

Substituting $x=\sinh t$, we get: $$I=\int_0^{\sinh^{-1}(1)} \frac{\cosh^2 t}{1+\sinh t}dt$$ This integral can be dealt with using the substitution $t=\log u$: $$I=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u+1/u)^2 du}{u(2+u-1/u)}=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^2+1)^2 du}{u^2(2u+u^2-1)}= \\ = \frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^4+2u^2+1) du}{u^2((...


2

The top radius $r$ of the cone is proportional to $h$: We have $r(t)=c\>h(t)$ for some constant $c$. Therefore $V(t)=c\> h^3(t)$ with some other $c$, or $h(t)=c\> V^{1/3}(t)$. This implies $${dh\over dt}=c\> V^{-2/3}(t) V'(t)=c\> V^{-2/3}(t)\ ,$$ since $V'(t)$ is constant. Now $t\mapsto V(t)$ is linear; hence $t\mapsto V^{-2/3}(t)$ is a "root ...


1

As is, I think that your integral $I$ is not ensured to be convergent. Take $f$ to be piecewise affine, with $f(0)=0$ and for all $n \geq 10$, $$f(n)=\frac{1}{n\ln{n}},$$ $$f\left(n \pm \frac{1}{\ln{n}}\right)=0.$$ Take $t_n=n+10$. Then $$\infty=\sum_n{f(t_{n-1})(t_n-t_{n-1})} \leq \sum_n{\int_{t_{n-1}}^{t_n}{\|f-f(t_{n-1})\| +\|f\|}} \leq I + \int_{\...


1

Why don't you just compute the integrals $$\int_{1}^{x} \frac{dt}{t}=\log(x)$$ $$\int_{\frac{1}{4}}^x \frac{dt}{4}=\frac{1}{4} \left(x-\frac{1}{4}\right)$$ lead to the equation $$\log(x)-\frac{1}{4} \left(x-\frac{1}{4}\right)=0$$ the solutions of which being given in terms of Lambert function. Skipping the intermediate steps (which are well described in the ...


1

Note that we got $$a^2+a+1=\frac{a^3-1}{a-1}$$ Thus \begin{align*} \frac13\log(a-1)-\frac16\log(a^2+a+1)&=\frac16\left(2\log(a-1)-\log\left(\frac{a^3-1}{a-1}\right)\right)\\ &=\frac16\log\left(\frac{(a-1)^3}{a^3-1}\right)\\ &=\frac16\log\left(\frac{a^3-3a^3+3a-1}{a^3-1}\right) \end{align*} As the logarithm is a continuous function we may ...


1

$$\int \csc(2x)dx=\int \frac{1}{\sin(2x)}dx=\int \frac{1}{2\sin(x) \cos(x)}dx=$$ $$\int \frac{\sin(x)}{2\sin^2(x) \cos(x)}dx=$$ $$\int \frac{\sin(x)}{2(1-\cos^2(x)) \cos(x)}dx$$ The substitution $u=\cos(x)$, $du=-\sin(x)dx$ changes the integral into $$\int \frac{du}{2u(u^2-1)}$$ which is solved by partial fractions.


1

$$\int \csc(2x)dx=\int \sec\left(\frac{\pi}{2}-2x\right)dx$$ Now, taking $u=\frac{\pi}{2}-2x$ we get $$-\frac{1}{2} \int \sec(u)du$$ and recall that, this is equal to $$-\frac{1}{2} \log \big| \sec(u)+\tan(u)\big|+c$$ Returning the variables $$-\frac{1}{2} \log \left| \sec\left(\frac{\pi}{2}-2x\right)+\tan\left(\frac{\pi}{2}-2x\right)\right|+c=-\frac{1}{2}\...


1

The conditions $x=0$ and $y=0$ represent the $y-$axis and $x-$axis respectively. We will first consider the limits at different points on the $x$ axis, excluding the point $(0,0)$: $$\lim_{(x,y)\to(a,0)} \left[x\cdot\sin\left(\frac{1}{y}\right) + y\cdot \cos\left(\frac{1}{x}\right)\right]$$ for $a\neq0$. If we approach the point $(a,0)$ along a line $x-a=ky$...


1

Note that, if $x,y\neq0$,\begin{align}\bigl\lvert f(x,y)\bigr\rvert&=\left\lvert x\sin\left(\frac1y\right)+y\cos\left(\frac1x\right)\right\rvert\\&\leqslant\lvert x\rvert+\lvert y\rvert\\&\leqslant2\sqrt{x^2+y^2}.\end{align}Besides, the inequality $\bigl\lvert f(x,y)\bigr\rvert\leqslant2\sqrt{x^2+y^2}$ also holds if $x=0$ or $y=0$. So, given $\...


1

Here is an elementary proof that $\dfrac{\sqrt{2}}{3n} \lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1) \lt \dfrac{2\sqrt{2}}{n} $. $\begin{array}\\ \sqrt{m+1}-\sqrt{m} &=(\sqrt{m+1}-\sqrt{m})\dfrac{\sqrt{m+1}+\sqrt{m}}{\sqrt{m+1}+\sqrt{m}}\\ &=\dfrac{1}{\sqrt{m+1}+\sqrt{m}}\\ &>\dfrac{1}{2\sqrt{m+1}}\\ \text{and}\\ \sqrt{m+1}-\sqrt{m} &<\...


1

This looks good. Just one comment. It's possible that the statement of the Intermediate Value Theorem you are invoking only applies to bounded intervals. In this case, you cannot invoke the theorem on the intervals $(-\infty,x_{0})$ and $(x_{0},+\infty)$. However, since you know $f(x)\to+\infty$ as $x\to-\infty$ and as $x\to+\infty$, you know that there are ...


1

It follows from the simple factorization property $$ \frac{\partial^2}{\partial x\partial y} f(x)g(y)=f'(x)g'(y)\ . $$


1

There is an error in the computation of the Jacobian: it is equal to $\frac r{2\sqrt3}$.


1

This is not a full answer, but maybe you can utilise the following result $$ \cos(2x) = 1 - 2\sin^2x\implies 1 - \cos(x) = 2\sin^2 \left(\frac{x}{2}\right) $$ we then have an integral $$ \int_0^\infty \frac{1-\cos x}{x^\alpha}dx = \frac{2}{2^\alpha}\int_0^\infty \frac{\sin^2 \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^\alpha}dx $$ then we have $$ \...


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