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8

Hint : Simplify it to $$\int \frac{1}{x^{2015}\left(1-\frac{1}{x^{2014}}\right)} \,dx$$ Then substitute $1-\frac{1}{x^{2014}}=t$


5

By setting $t=\sinh u$ we get: $$ I=\int_{0}^{\log(1+\sqrt{2})}\sinh^4(u)\cosh^2(u)\,du=\\\frac{1}{64}\int_{0}^{\log(1+\sqrt{2})}\left(4+e^{-6 u}-2 e^{-4 u}-e^{-2 u}-e^{2 u}-2 e^{4 u}+e^{6 u}\right)\,du $$ that is not a difficult integral to compute, just a tedious one. It boils down to: $$ I = ...


4

Write $\sin 2x = 2 \sin x \cos x$. The substitution $u = \sin x$ will yield $$2 \int e^u u \, du$$ which you can integrate by parts.


4

For both integrals you can use substitution. For the top: Let $u=\cos x$, so $du=-\sin x dx$. When $x=0$ then $u=1$. When $x=\frac{\pi}{2}$ then $u=0$. $$\int_0^{\frac{\pi}{2}} cos^3x \sin x dx=-\int_1^0 u^3 du=-\frac14u^4\bigg|_1^0=\frac14$$ $$\int_0 For the bottom: Let $u=\sin x$ so $du=\cos x dx$ When $x=0$ then $u=0$. When $x=\frac{\pi}{2}$ ...


4

$${\frac{\ln(kx)}{k}+C = \frac{\ln(k)}{k}+\frac{\ln(x)}{k}+C = \frac{\ln(x)}{k} + \tilde{C}}$$ So they're the same, they just differ by a constant


3

Without evaluating this integral: Note $1\le 2-\sin x\le 2$ on $\;[0,\frac\pi2]$, hence $\;\dfrac{x^2}2\le \dfrac{x^2}{2-\sin x}\le x^2 $, and by the positivity of the integral: $$\frac{\pi^3}{48}=\int_0^\tfrac\pi2\dfrac{x^2}2\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2\dfrac{x^2}{2-\sin x}\,\mathrm d\mkern1mu x\le \int_0^\tfrac\pi2 x^2\,\mathrm d\mkern1mu ...


3

With the fundamental theorem of calculus, since $\;\sin t^2\;$ is continuous everywhere, you have $$\int_{\sqrt x}^0\sin t^2\;dt=F(0)-F(\sqrt x)\;,\;\;\text{with}\;\;F'(t)=\sin t^2$$ Thus, to differentiate the above, just apply the chain rule and use the fact that $\;F\;$ is a primitive of the integrand: $$\left(\int_{\sqrt x}^0\sin ...


3

Since $2-\sin x\in[1,2]$ for $x\in [0,\frac\pi2]$, we have $$ \frac12\int_0^{\pi/2}x^2\,dx \le \int_0^{\pi/2}\frac{x^2}{2-\sin x}\,dx \le \int_0^{\pi/2}x^2\,dx. $$ Computing the integral of $x^2$ gives you the desired inequalities.


3

Your formula $\dfrac{x^2-x}2$ works as an approximation, but it fails as an actual formula. For example, the actual area "under" the floor function (below it and above the $x$-axis) between $x=0$ and $x=\frac 12$ is zero, but your formula gives $$\left[\frac{x^2-x}2\right]_0^{1/2}=-\frac 18$$ Are you trying for an approximation or for an actual formula? ...


2

You can use residue theorem. $$e^{1/z}=1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+\dots$$ $$\frac{e^{1/z}}{z^2}=\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{2z^4}+\frac{1}{6z^5}+\dots$$ So the residue of $\frac{e^{1/z}}{z^2}$ in $z=0$ is $0$. So $\int_{\gamma} \frac{e^{\frac{1}{z}}}{z^{2}} \; dz =0$.


2

Both are fine, they differ by a constant. and actually $$\int \frac{1}{x} dx = \ln |x| $$


2

We want to compute: $$ I =\int_{0}^{+\infty}\frac{1-e^{-x}}{x(e^x+1)(e^{2x}+1)}\,dx =\int_{0}^{+\infty}\left(\frac{1}{e^{2x}+1}-\frac{1}{e^{2x}+e^{x}}\right)\frac{dx}{x}$$ that is the same as computing $$ J = \int_{0}^{1}\frac{t^2(1-t)^2}{(1-t^4)\log t}\,dt.$$ Since $\frac{t^2(1-t)^2}{1-t^4}=\sum_{k\geq 1}\left(t^{4k-2}-2t^{4k-1}+t^{4k}\right)$ and ...


2

By parts works perfectly, $$\int2\sin(x)\left(\cos(x)e^{\sin(x)}\right)dx=2\sin(x)e^{\sin(x)}-2\int\cos(x)e^{\sin(x)}dx\\ =2\sin(x)e^{\sin(x)}-2e^{\sin(x)}$$


2

Given that the population has a logistic form, at long times (large t), the population reaches a steady state. Thus $dP/dt = 0$ as t approaches infinity. Solving for P yields $\pm 112$


1

You are good except for your last step; $\tan(\theta)$ is not $x^2$, and in particular $\csc$ is not $1/\sec$. One way to identify what $\tan(\theta)$ is in terms of $x$ is to consider the triangle that justifies the substitution $x=\sec(\theta)$ in the first place. This has a leg of $\sqrt{x^2-1}$, which means the hypotenuse is $x$ and the other leg is ...


1

Hint: This is a autonomous differential equation. With $p=0$ and $p=\pm \sqrt{224\cdot 56}=\pm112$ as trivial constant solutions. Draw an approximate plot (look up how to draw solutions for autonomous differential equations) of the solutions and see what happens for different initial values for $t \to \infty$.


1

Do the usual and draw the region, and check that (fill in details) $$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\;\;,\;\;\frac\pi4\le t\le \frac\pi2\;,\;\;0\le r\le 2\sqrt2$$ so your integral becomes (don't forget the Jacobian!) ...


1

$$\sin(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)!} $$ hence: $$\sin(x^3)=\sum_{n\geq 0}\frac{(-1)^n x^{6n+3}}{(2n+1)!}$$ and: $$\frac{\sin(x^3)}{x^3}=\sum_{n\geq 0}\frac{(-1)^n x^{6n}}{(2n+1)!}$$ so: $$ \int\frac{\sin(x^3)}{x^3}\,dx = C+\sum_{n\geq 0}\frac{(-1)^n x^{6n+1}}{(6n+1)\cdot(2n+1)!}.$$ Also notice that: $$ ...


1

Substituting $x=\frac{t^2-1}{2t}$, we get $dx= \frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$ Plugging this in our integral gives: $$\int \frac{\frac{1}{2}(1+\frac{1}{t^2})dt}{\sqrt{1+\left(\frac{t^2-1}{2t}\right)^2}}$$ $$\text {(Skipped some algebra)}$$ $$=\int \frac{(t^2+1)dt}{t^2\sqrt{\left(\frac{t}{2}+\frac{1}{2t}\right)^2}}$$ $$\text {(Skipped some ...


1

For the first part, since $g$ is of compact support, it is bounded. Hence $\|g\|_{\infty} < \infty$, so $$\int |f_ng| = \int |f_n| |g| \le \int |f_n| \|g\|_{\infty} = \|g\|_{\infty} \int |f_n| \to 0$$ For the second, note that: $\overline{C^{0}_{cpt}} = L^1$ (wrt $\|\cdot \|_1$), so for $g \in L^1$ and $\epsilon > 0$, there exists a continuous ...


1

Starting with the integral $$I =\int_0^1 t^4\sqrt{1+t^2}\,\text{d}t,$$ use the substitution $t = \tan(\theta)$, $\text{d}t = \sec^2(\theta)\,\text{d}\theta$; when $t = 0$, $\theta = 0$, when $t=1$, $\theta =\pi/4$. This yields the integral $$I = \int_0^{\pi/4} \tan^4(\theta)\sqrt{1+\tan^2(\theta)}\sec^2(\theta)\,\text{d}\theta = \int_0^{\pi/4} ...


1

Assume that $f$ is continuous and that $\phi$, $g$ are differentiable. Suppose $$H(x) = \int_0^x f(s) \, ds.$$ The fundamental theorem of calculus tells you that $H'(x) = f(x)$. Suppose that $$M(x) = \int_0^{\phi(x)} f(s) \, ds.$$ Then $M(x) = H(\phi(x))$ so that the chain rule tells you $$M'(x) = H'(\phi(x)) \phi'(x) = f(\phi(x)) \phi'(x).$$ The ...


1

Since the only pole of the function within the enclosed domain by the given path is zero, we get $$\oint_{|z|=1/2}\frac{\frac{e^{1-z}}{1-z}dz}{z^3}=\left.\frac{2\pi i}{2!}\left(\frac{e^{1-z}}{1-z}\right)''\right|_{z=0}=e\pi i$$ Another way: with a few summands of the power (Laurent) series ( observe we want close to zero so $\;|z|<1\;$) : ...


1

Maybe you remember that there is a right triangle with sides $3$, $4$, $5$. This then implies that $$u:={4\over5}x-{3\over5}y,\qquad v:={3\over5}x+{4\over5}y$$ are orthonormal coordinates in the plane. In terms of these coordinates the given equation reads $$\bigl\lfloor |v|\bigr\rfloor+\bigl\lfloor |u|\bigr\rfloor=3\ .\tag{1}$$ Draw an $(u,v)$-plane with ...


1

We can separate it into cases : $$\left\lfloor\frac{|3x+4y|}{5}\right\rfloor+\left\lfloor\frac{|4x-3y|}{5}\right\rfloor=3$$ $$\iff \left(\left\lfloor\frac{|3x+4y|}{5}\right\rfloor,\left\lfloor\frac{|4x-3y|}{5}\right\rfloor\right)=(0,3),(1,2),(2,1),(3,0)$$ Now note that $$\left\lfloor Z\right\rfloor=a\iff a\le Z\lt a+1$$ and that ...


1

You can take the derivative of an integral, even though you can't directly integrate it. let $I = \int \sin t^2 dt$ The value that you care about is $$R = I(0) - I(\sqrt x)$$ since we are asked to evaluate the integral from $\sqrt x$ to $0$. Now, differentiating $R$ by $dx$, we get the expression $$\frac{dR}{dx} = \frac{d I(0)}{dx} - \frac{dI(\sqrt ...


1

Notice that $\frac{1}{2} ~\leq ~\frac{1}{2-\sin(x)}~\leq ~1$ for $x \in [0, \pi/2]$.


1

Hint: For $0 \leq x \leq \frac{\pi}{2}$, $0 \leq \sin(x) \leq 1$.


1

Rory Daulton already gave a complete answer, but here's a slightly different way of coming to an equivalent conclusion. Compare the region of the plane between the $x$ axis and the graph of the function $f_1(x) = x$. Now do the same for the function $f_2(x) = \lfloor x \rfloor$. The difference between the two regions is a sequence of right triangles that ...



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