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5

This is true for $a\in \mathbb Z$ too. $-(\pi-x)a$ is real, so $-i(\pi-x)a$ is purely imaginary. It is well known that $$ e^{it}=\cos t + i\sin t \qquad\qquad\text{when }t\in\mathbb R$$ so its modulus is $\sqrt{\sin^2 t+\cos^2 t} = \sqrt 1 = 1$.


4

The image below might help to explain where this term comes from:


4

We have: $$I=\int_{0}^{\pi/2}e^{-\sin x}\,dx = \int_{0}^{1}\frac{e^{-t}}{\sqrt{1-t^2}}\,dt $$ and since: $$ e^{-t}=\sum_{k\geq 0}\frac{(-1)^k t^k}{k!},\qquad \int_{0}^{1}\frac{t^k}{\sqrt{1-t^2}}\,dt =\int_{0}^{\pi/2}\sin^k\theta\,d\theta=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{2\,\Gamma\left(\frac{k}{2}+1\right)}$$ it follows ...


3

Depict carefully the path of integration: it is a semicircle in the upper half plane with a bulge at $z=0$ and a keyhole around $z=i$. This gives that you have to compute the residues of $f(z)=\frac{e^{iz}}{z(z^2+1)^2}$ at $z=0$ and $z=i$, but to consider only half the residue at $z=0$: ...


3

Another way to evaluate this integral is to use Parseval's theorem, which states that for functions $f$ and $g$ with respective Fourier transforms $F$ and $G$, we have $$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$ Here $f(x) = \sin{x}/x$ and $g(x) = (1+x^2)^{-2}$. Thus, $F(k) = \pi$ when $k \in ...


3

The current approach seems reasonably efficient to me: From this point, if you expand out the dot product, you'll get just three terms, each of them manageable integrals involving sines and cosines. You can handle parts of the integrals quickly using the fact that the interval has length $2\pi$, which is the period of the $\sin$ and $\cos$ functions that ...


3

Rewrite $I_n$ in this way: $$ I_n = -\frac{1}{3}\int x^{n-1} \frac{3}{2}(1-x^2)^\frac{1}{2} (-2x) \, dx$$ and note that derivative of $(1-x^2)^{\frac{3}{2}}$ is $\frac{3}{2}(1-x^2)^\frac{1}{2} (-2x)$. Then integrate by parts and rearrange.


2

$|\sin x|\sin(nx)$ is an odd function, so its integral on the interval $[-\pi,\pi]$ is equal to $0$. On the other hand $|\sin x|\cos(nx)$ is even. Its integral on $[-\pi,\pi]$ is twice the integral on $[0,\pi]$. The last integral should be from $0$ to $\pi$.


2

We have $$\int_0^\infty\frac{\sin x}{x}dx=\int_0^\infty\int_0^\infty e^{-xt}\sin (x)dtdx=\int_0^\infty\operatorname{Im}\int_0^\infty e^{x(i-t)}dxdt\\=\int_0^\infty\frac1{1+t^2}dt=\arctan t\Bigg|_0^{\infty}=\frac\pi2$$


2

$f$ is a continuous function on the compact $[a,b]$ so it's bounded and then there's $m,M$ such that $$m\le f(x)\le M,\quad\forall x\in[a,b]$$ so $$m=\frac1{b-a}\int_a^b mds\le \frac1{b-a}\int_a^b f(s)ds\le \frac1{b-a}\int_a^b Mds=M$$ and the result follows by applying the intermediate value theorem.


2

As a rule of thumb, it's better to use $$\frac12\int_{\partial B}(-y\,dx+x\,dy)$$ whenever trig functions are involved. That will give a simpler expression once you use standard trig identities. Comment: It's more straight forward in this particular case to do the area integral directly in polar coordinates: $$\int_0^{2\pi}\int_0^{1+\cos\theta} ...


2

...Hint: $\dfrac{1}{\sqrt{t}+\sin t} \leq \dfrac{1}{\sqrt{t}}$


2

This is a very challenging integral. I don't feel the result is very intuitive, but I learned how to solve it in a PDE's class years ago. First, Euler's identity: $$\int^{2\pi}_{0} e^{ia \cos{\theta}}d\theta = \int^{2\pi}_{0} \cos(a \cos{\theta})d\theta+\int^{2\pi}_{0}i\sin(a \cos{\theta})d\theta $$ First, let's solve $\int^{2\pi}_{0} \cos(a ...


2

Hint: Try breaking the numerator into $(x^2+2x+2)'=2x+2$ and a constant and trying to complete the square in second term.


1

Split the integral into to parts: from 0 to $\pi/2$ and to $\pi/2$ to $\pi$, the second is equal to the first (you can prove it by the following substitution: $t = \pi - \theta$). To calculate the first integral you can divide all by $(cos(\theta))^2$ and then make the substitution $x = tan(x)$. From there use simple fractions and you're done. You could ...


1

To make the manipulations simpler we first substitute $x-2=u$ and this means $du=dv$ and we are left with $$\int\arctan{(x-2)}dx=\int\arctan{u}du$$ Now we integrate by parts $v=\arctan{u}$ and $du$. This yields $dv=\frac{1}{u^2+1}du$ and $u$. We have therefore $$\int\arctan{u}du=u\arctan{u}-\int\frac{u}{u^2+1}du$$ and with a substitution $w=u^2+1$ we are ...


1

The integral equals $$ \mathrm{B}(1/2,1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{\sqrt \pi \sqrt \pi}{1} = \pi. $$


1

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1

Let $S$ be your integral, so \begin{align} S = \int e^{-x}\left(x+y\right) \,\mathrm{d}x&= \int \frac{x}{e^x}\,\mathrm{d}x + y \int \frac{1}{e^x} \,\mathrm{d}x\\ \end{align} Now let $u=x$, with $\mathrm{d}u=\mathrm{d}x$. Let $v=-e^{-x}$. Then $\mathrm{d}v = e^{-x} \,\mathrm{d}x$. This means we can write $S$ as \begin{align} S &= \int u ...


1

Hint Let $x=\dfrac{\pi}{2}+t$,then $$\sin^n{x}\sin{(n+2)x}=\cos^{n}{t}\sin{[(n+2)(\frac{\pi}{2}+t)]}$$ is odd function $$I=\int_{0}^{\pi}\sin^n{x}\sin{(n+2)x}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{n}{t}\sin{[(n+2)(\frac{\pi}{2}+t)]}dt=0$$


1

You forgot the square root $$x=\frac{1}{2} \tan \theta \implies dx= \frac{1}{2} \sec^2 \theta d\theta $$ The integral becomes: $$\int \frac{1}{\sqrt{1+tan^2 \theta}} \cdot \frac{1}{2} \sec^2 \theta d\theta$$ $$\Large{\text{Not:}}$$ $$\bbox[8pt, border: crimson 4pt solid]{\int \frac{1}{1+tan^2 \theta} \cdot \frac{1}{2} \sec^2 \theta d\theta}$$


1

What happened to the square root? $\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=|\sec\theta|$. So you have $$\frac12\int\frac{1}{|\sec\theta|}\sec^2\theta\,d\theta=\frac12\int|\sec\theta|\,d\theta$$ In the original substitution $2x=\tan\theta$, $\theta$ is in a range where $\sec\theta$ is positive, so actually we just have ...


1

If $x = \frac{1}{2} \tan \theta$, then $\theta = \tan^{-1} 2x$. But you have a more important problem: the integrand you wrote is $$\frac{1}{\sqrt{1+4x^2}},$$ whereas what you integrated was $$\frac{1}{1+4x^2},$$ because you forgot the square root. You should have instead written $$\int \frac{dx}{\sqrt{1+4x^2}} = \frac{1}{2} \int \sec \theta \, d\theta.$$ ...


1

Here are some facts you can use: $\displaystyle\int_1^{\sqrt{x}}\frac{1}{u}du=\big[\ln u\big]_1^{\sqrt{x}}$, $\;\;\ln(\sqrt{x})=\frac{1}{2}\ln x$, $\;\;\displaystyle\frac{\sqrt{x}-1}{x}<\frac{1}{\sqrt{x}}, \;\;\ln\left(\frac{1}{t}\right)=-\ln t$


1

Big Hint: We know by the fundamental theorem of calculus that $$\int_{1}^{\sqrt{u}} \frac{du}{u} = \frac{ln(x)}{2}.$$ However, $$\int_{1}^{\sqrt{x}} \frac{du}{u} < \sqrt{x} -1.$$ Thus, we can say $$0<\frac{ln(x)}{2} < \sqrt{x} -1 \text{ or } 0<lnx<2(\sqrt{x} -1).$$ Now, if we divide by $x$ and apply Squeeze Theorem, one will quickly ...


1

Here are the steps $$\int\frac{8}{x^2+49}dx=\frac{8}{49} \int\frac{1}{\frac{x^2}{49}+1}dx $$ Let $u=\frac{x}{7}$, then $du=\frac{1}{7}dx$. So now $$ \frac{8}{7} \int\frac{1}{u^2+1}du= \frac{8}{7} \arctan u+C $$ $$ = \frac{8}{7} \arctan\left(\frac{x}{7}\right)+C $$



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