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1

Your error is that : $$\int_{0}^{\pi}\sqrt{\left(\sin\left(\frac{x}{2}\right)-\frac{1}{2}\right)^{2}}\mathrm{d}x \neq \int_{0}^{\pi}\left(\sin\left(\frac{x}{2}\right)-\frac{1}{2}\right)\mathrm{d}x \\$$ but instead : \begin{align}\int_{0}^{\pi}\sqrt{\left(\sin\left(\frac{x}{2}\right)-\frac{1}{2}\right)^{2}}\mathrm{d}x & = ...


0

You lost the modulus, when you erase the square root within the integral.


1

Your third equality sign: you forgot that $\sqrt{x^{2}}=|x|$. Therefore, you should take an absolute value and then split the integral into positive and negative part. The $\sin (x/2)-1/2$ function changes the sign between $0$ and $\pi$.


2

$$\begin{align}\int_0^1 dx \frac{x^4 (1-x)^4}{1+x^2} &= \int_0^1 dx \frac{x^4-4 x^5 + 6 x^6 - 4 x^7 + x^8}{1+x^2}\\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{4+2 k}-4 x^{5+2 k}+6 x^{6+2 k}-4 x^{7+2 k}+x^{8+2 k}\right )\\&= \sum_{k=0}^{\infty} (-1)^k \left (\frac1{5+2 k}-\frac{4}{6+2 k}+\frac{6}{7+2 k}-\frac{4}{8+2 k}+\frac1{9+2 k} ...


0

Even if $f(a)=f(b)$ there is no difference than usual way of integrating. Try to get antiderivative. And $\int_0^k f(x)dx = 2\int _0^{k/2}f(x)dx$ can be only used if $f(x)=f(k-x)$ for all $x$. Since it is symmetric to $k/2$, $\int_0^{k/2}f(x)dx$ is equal to $\int _{k/2}^{k}f(x)dx$. So following is true. $$\int_0^k f(x)dx = \int_0^{k/2}f(x)dx +\int ...


2

Literally all you have to do is expand it out, divide, and integrate. $$ x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4 $$ Now use long division: $$ x^4(1-x)^4 = x^6(x^2+1)- 4x^5(x^2+1) + 5x^4(x^2 + 1) - 4x^2(x^2+1)+4(x^2+1) -4 \\ \frac{x^4(1-x)^4}{1+x^2} = (x^6 -4x^5 + 5x^4 - 4x^2 + 4) - \frac{4}{x^2+1} \\ $$ We integrate this: $$ \int_0^1 \! (x^6 -4x^5 + ...


3

Hint: First, prove that $I_n(t)=\displaystyle\int_{-\infty}^\infty\frac{\cos(tx)}{x^2+n^2}dx=\frac\pi n\cdot e^{-nt}$ . Then express your integral in terms of $I'_a(1)$ and $I'_b(1)$, since $\dfrac{d}{dt}\cos(tx)=-x\sin(tx)$, which, for $t=1$, becomes the numerator of our integrand.


0

It may be good to try partial fractions. The result it: $$ \int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)} dx = \frac{\pi}{a^2-b^2}\left(\sinh(a)-\cosh(a)+\cosh(b)-\sinh(b) \right) $$


3

Any ideas what's going wrong here? Yes. $\displaystyle\int_{-\infty}^\infty e^{-x^{2k}}dx$ cannot be computed by expanding $e^t$ into its Taylor series, and then switching the order of summation and integration. It just doesn't work. All you get is an alternating sum of $\pm\infty$, which is undetermined.


0

So take $3\int_{-1}^0(x^3-x)dx$, let $x=-y\Rightarrow dx=-dy$, when $x=0, y=0$, and when $x=-1, y=1$, so we get: $3\int_{-1}^0(x^3-x)dx=3\int_{1}^0((-y)^3-(-y))(-dy)=3\int_0^1(y-y^3)dy=-3\int_0^1(y^3-y)dy$ $=-3\int_0^1(x^3-x)dx$ Thus: $3\int_{-1}^0(x^3-x)dx-\int_0^1(x^3-x)dx=-3\int_0^1(x^3-x)dx-3\int_0^1(x^3-x)dx=-6\int_0^1(x^3-x)dx$ To find the other ...


2

Hint If $f$ is an odd function then (using the change of variable $t=-x$) $$\int_0^a f(x)dx=-\int_{-a}^0f(x)dx$$


0

We can use the parametric integral to evaluate the integral. Let $$ I(\alpha)=\int_0^\infty\frac{\ln(1+\alpha x)}{(1+x^2)(1+x^3)}dx. $$ Then it is easy to see \begin{eqnarray*} I'(\alpha)&=&\int_0^\infty\frac{x}{(1+\alpha x)(1+x^2)(1+x^3)}dx\\ ...


1

There is another way to solve. Let $$ I(\alpha,\beta)=\int_0^\infty\frac{\ln(1+\alpha x)\ln(1+\beta x)}{x^{\frac{3}{2}}}dx. $$ Note $I=I(1,1)$. It is easy to see that $$ \frac{\partial^2I(\alpha,\beta)}{\partial\alpha\partial\beta}=\int_0^\infty\frac{\sqrt{x}}{(1+\alpha x)(1+\beta x)}dx=\frac{\pi}{\alpha\sqrt{\beta}+\sqrt{\alpha}\beta} $$ and hence $$ ...


0

One way to look at this is that $f$ being analytic means it is holomorphic, and thus satisfies the cauchy reimann equations. Now consider laplace's equation in $\mathbb { R }^2$: $-\Delta f=0\Rightarrow -(\frac{\partial^2}{\partial x^2}+\frac {\partial^2}{\partial y^2})f=0\Rightarrow - (\frac{\partial}{\partial x}+i\frac {\partial}{\partial ...


2

Hint: If we let $z=z_0+Re^{i\theta}$, then $$ \frac1{2\pi}\int_0^{2\pi}f(z_0+Re^{i\theta})\,\mathrm{d}\theta =\frac1{2\pi i}\oint f(z)\frac{\mathrm{d}z}{z-z_0} $$ where the path is the circle of radius $R$ counterclockwise around $z_0$. Next, look for the singularities of the integrand then use Cauchy's Integral Formula.


11

Let $z=e^{i x}$; then $dx = -i dz/z$ and the integral is equal to $$-i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$ By the residue theorem, the integral is then equal to $i 2 \pi$ times the sum of the residues at each pole inside the unit circle. The residue at each pole $z_k$ is equal to $$-i 8 \frac{z_k^3}{8 z_k^7 + 24 z_k^3} = -i \frac1{z_k^4+3}$$ ...


7

We have $\displaystyle\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$ $\displaystyle=\frac{2-\sin^22x}2=\frac{2(1+\tan^22x)-\tan^22x}{2\sec^22x}=\frac{\tan^22x+2}{2\sec^22x}$ $$\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{2\sec^22x}{\tan^22x+2}dx$$ Setting $\tan2x=u,$ ...


3

Note that $\tan 2x$ is discontinuous at $\frac{\pi}4$ and some other values which can be easily found. You have to break the integral from $0$ to $\frac{\pi}4$ and so on.


0

All the derivatives $f^{(n)}$ with $n$ odd are $\pm\sin$, and $$\int_0^{2\pi}\sin(kx)\,dx=\left.-{\cos kx\over k}\right]_{x=0}^{x= 2\pi}=0.$$ The calculations for $n$ even are similar.


1

Assume $a \gt 0$. Ignore the $\operatorname{sgn}^2{x}$, as that is identically $1$ almost everywhere, except at $x=0$ (which can be ignored as it is measure zero). Thus you want to find $$-i \int_{-\infty}^{\infty} dx \frac{x \, e^{i x}}{1+a x^2}$$ Then consider the integral $$-i \oint_C dz \frac{z \, e^{i z}}{1+a z^2}$$ where $C$ is a semicircle of ...


0

If the answers don't look like yours, there's chance that you've done something wrong. The integral of $\ln x$ is not $\frac{1}{2} \ln^2 x$. It's evident when you take the derivative: $$ \frac{d}{dx} \left( \frac{1}{2} \ln^2 x \right) = \frac{\ln x}{x} $$ For the $\ln$ function, try integration by parts. Also, use a substitution first, don't think you can ...


2

Let $f(x) = x + x^7$, and $g(t)$ its inverse function on $[0,\infty)$. Then $$\int_0^x \sin(x + x^7)\ dx = \int_0^{f(x)} \sin(t) g'(t)\ dt$$ It can be shown that as $t \to \infty$, $$g(t) = t^{1/7} - \dfrac{1}{7} t^{-5/7} + O(t^{-11/7})$$ and $$g'(t) = \dfrac{1}{7 g(t)^6 + 1} = \dfrac{1}{7} t^{-6/7} + O(t^{-12/7})$$ Now $\int_1^\infty |\sin(t)| ...


4

Let $$I = \int_{\ln(4)}^{\ln(6)} e^x \ln(e^{2x}-4)dx$$ Now set $t=e^x$ and we get $$I = \int_4^6 \ln(t^2-4) dt = \int_4^6 \ln(t+2) dt + \int_4^6 \ln(t-2) dt = \int_6^8 \ln(t)dt + \int_2^4 \ln(t)dt$$ I trust you can finish it from here.


1

Doing your procedure, one should obtain: $$ \int_{u=0}^{u=\frac{\pi}{2}} 8u^3 \cos(u) = 8\left(u(u^2-6)\sin(u)+3(u^2-2)\cos(u)\right) \left. \right|_{0}^{\frac{\pi}{2}}=48-24\pi + \pi^3 $$


1

$\sinh^{-2} x$ is an odd function: $$\sinh^{-1}(x):=\ln(x+\sqrt{x^2+1})\\ \implies \sinh^{-1}(-x)=\ln(-x+\sqrt{x^2+1})=\ln(-(x-\sqrt{x^2+1}))=\ln\left(\dfrac{x-\sqrt{1+x^2}}{x^2-1-x^2}\right)=\\ \ln\left(\dfrac{1}{x-\sqrt{1+x^2}}\dfrac{x+\sqrt{1+x^2}}{x+\sqrt{1+x^2}}\right)=\ln\left(\dfrac{1}{x-\sqrt{1+x^2}}\right)=-\ln(x+\sqrt{x^2+1})\\ \implies ...


4

$\dfrac{d}{dx}\cos x=-\sin x$, hence, $$\int^\pi_0\sin xdx=-\left.\cos x\right|^\pi_0=1-(-1)=2$$


1

In fact the last part of your second equation should be:$$\ldots=1-(-1)=2$$ :)


1

Hint: The function $$\frac{1}{64 + x^6}$$ is even, and the product of an even function and an odd one is odd. This reduces the problem to considering only one of the terms.


3

We have $$\frac{x^5+x^2+4x+\sin(x)}{64+x^6}=\underbrace{\frac{x^5+4x+\sin(x)}{64+x^6}}_{=f(x)}+\underbrace{\frac{x^2}{64+x^6}}_{=g(x)}=f(x)+g(x)$$ Now $f(-x)=-f(x)$, so $f$ is an odd function. Also $g(x)=g(-x)$, so $g$ is an even function. That is $$\int_{-2}^2 f(x)=0$$ and $$\int_{-2}^2 g(x)= 2\int_0^2 g(x) dx$$ $g$ is easy to integrate: a primitive ...


3

$$\int^2_{-2}\dfrac{x^5+x^2+4 x+\sin(x)}{64+x^6} dx=\int^2_{-2}\dfrac{x^5}{64+x^6} dx+\int^2_{-2}\dfrac{x^2}{64+x^6} dx+\\ \int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\ \left.\dfrac{1}{6}\ln|64+x^6|\right|^2_{-2}+\int^2_{-2}\dfrac{x^2}{1+\left(\frac{x^3}{8}\right)^2} dx+\int^2_{-2}\dfrac{4 x}{64+x^6} ...


1

The length element for a polar curve $\theta\mapsto r(\theta)$ is $\mathrm{d}\ell=\sqrt{r(\theta)^2+r'(\theta)^2}\mathrm{d}\theta$. This fact is easy to prove as follows (and should be a known result): from $$\mathrm{d}(r\vec e_r)=(\mathrm{d}r)\vec e_r+r\mathrm{d}\vec e_r=r'(\theta)\vec e_r\mathrm{d}\theta+r(\theta)\vec e_\theta\mathrm{d}\theta$$ we obtain ...


0

You can also use the substitution $x-1=t^{2}$ which then you will have the derivative of arctan in the integral.


2

Your approach is fine. We get the integral as $$I = \int \dfrac{dx}{\sqrt{x-1}(x+4)}$$ Now let $x-1 = t^2$. We then get $$I = \int \dfrac{2tdt}{t(t^2+5)} = \int\dfrac{2dt}{t^2+5} = \dfrac2{\sqrt5}\arctan(t/\sqrt5) + \text{constant} = \dfrac2{\sqrt5}\arctan\left(\dfrac{\sqrt{x-1}}{\sqrt5} \right) + \text{constant}$$ Now I trust you can finish it off.


2

HINT: As $x$ lies $\displaystyle\in[6,16] \sqrt{(x+4)^2}=+(x+4)$ Set $\displaystyle\sqrt{x-1}=u\implies x=u^2+1$ Alternatively, Using Trigonometric substitution start with $\displaystyle\sqrt{x-1}=\tan\theta$ $\displaystyle\implies x=\sec^2\theta$ $$\int\frac1{\sqrt{x-1}(x+4)}dx=\int\frac{2\sec^2\theta\tan\theta}{\tan\theta(\sec^2\theta+4)}\ d\theta$$ ...


3

$S$ is term inside square root. $S_1=1+3 \sin^2 x$ $S_2=\sin^4 x+\cos^4 x-\sin^2 x \cos^2 x$ $=1-3\sin^2 x \cos^2 x$ Bring that 2 inside $=4-3\sin^2 2x$ Put $2x=t \implies 2dx=dt$ Limit will go to $4\pi$ but $dt/2$ shall counter it due to periodicity. $=4-3\sin^2 t$ $=1+3\cos^2t$ Plotting both functions in a rough graph will lead to believe that ...


1

I do not know how much this could help $$4 \sin ^2(t)+\cos ^2(t)-4 \left(\sin ^6(t)+\cos ^6(t)\right)=-\frac{3}{2} (\cos (2 t)+\cos (4 t))$$


0

Initial attempt: $4sin^{2} + cos^{2} = 3sin^{2}+1$ $sin^{6} + cos^{6} =sin^{4} -sin^{2}cos^{2} +cos^{4}$ Will get back to this later


2

$$\log(1+e^{2ix}) = \log(e^{i x}(e^{-ix}+ e^{i x})) = \log(e^{ix})+ \log(2 \cos x) = ix + \log(2 \cos x)$$ So $$ \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx = \int_{0}^{\pi /6} \Big( ix + \log(2 \cos x) \Big)^2 \ dx$$ Then equating the real parts on both sides of the equation and rearranging, $$ \int_{0}^{\pi /6} \log^{2}(2 \cos x) \ dx = \int_{0}^{\pi ...


0

This is really more of an appendix to Robert's answer. You have an integral dependent on six parameters, which we'll denote by $I_n(p,q;a,b,c)$. Some of the complexity of the integrand from the parameter dependence is entirely superficial and can be eliminated by transformations as simple as a change of scale. As you can see from the nasty cumbersome ...


0

Even for the case $n=1$ Maple gives a rather complicated result involving hypergeometric functions: $$ {\frac {{c}^{p}{a}^{-q+1}}{\Gamma \left( q \right) } \left( {c}^{-1-p }{a}^{1+p}{\mbox{$_2$F$_1$}(1,2+p;\,3-q+p;\,{\frac {a}{c}})}\Gamma \left( 2+p \right) \Gamma \left( -p+q-2 \right) +\pi \,{c}^{-q+1}{a} ^{q-1} \left( 1-{\frac {a}{c}} \right) ...


4

It turns out that the Fourier transform of $J_0^3$ can still be expressed in terms of complete elliptic integrals, but it's considerably more complicated than the formula for ${\cal FT}(J_0^2)$: for starters, it involves the periods of a curve $E$ defined over ${\bf C}$ but (except for a few special values of $\omega$) not over ${\bf R}$. Assume $|\omega| ...


0

$$\int_0^\pi\theta^2\ln^2\left(2\cosh\dfrac{\theta}{2}\right)d\theta$$ $$=\int_0^\pi\theta^2(\ln(e^\frac{\theta}{2}+e^{-\frac{\theta}{2}}))^2~d\theta$$ $$=\int_0^\pi\theta^2(\ln(e^\frac{\theta}{2}~(1+e^{-\theta})))^2~d\theta$$ $$=\int_0^\pi\theta^2\left(\dfrac{\theta}{2}+\ln(1+e^{-\theta})\right)^2~d\theta$$ ...


1

You should split the triple integral into two parts. $$V = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^{\frac{1}{cos \phi}} \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta+ \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_0^2 \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta $$ Note that the first one is the volume of a cone.


1

HINT: $$\int_0^1x^5\sqrt{1-x^2}\ dx=\int_0^1 x^4\sqrt{1-x^2}\ xdx$$ Set $\displaystyle 1-x^2=y^2\implies x\ dx=-y\ dy$ and $x^2=1-y^2$


0

Let $$I_n = \int_0^{\pi/2} \dfrac{\sin(2n+1)x}{\sin(x)}dx$$ We then have $$I_{n+1} - I_n = \int_0^{\pi/2} \dfrac{2 \sin(x)\cos(2(n+1)x)}{\sin(x)} dx = 2\int_0^{\pi/2} \cos(2(n+1)x) dx = 0$$ for $n \neq -1$. Hence, $$I_n = I_0 = \dfrac{\pi}2$$ Essentially, we are saying that $\dfrac{dI_n}{dn} = 0$, where the derivative is to be interpreted in discrete sense ...


2

The only way I found is based on the infinite Taylor expansion of the integrand. For the first terms, we have $$x^2\ln(\sinh x)\ln(\cosh x)=\frac{1}{2} x^4 \log (x)+x^6 \left(\frac{1}{12}-\frac{\log (x)}{12}\right)+x^8 \left(\frac{\log (x)}{45}-\frac{1}{60}\right)+x^{10} \left(\frac{197}{45360}-\frac{17 \log (x)}{2520}\right)+x^{12} \left(\frac{31 \log ...


1

Try doing what you did to find the circumference of a circle. It's like you approximated the circumference with a bunch of vertical line segments. But no matter how many you make and how small you make them, they still just have a total lenght of $4R$. They're not doing a good job of approximating the curve. To do that, you need to make the ones closer to ...


3

$$ \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n} $$ $$ = \int_0^\infty dy_1 \dots \int_0^\infty dy_n \int_{s=0}^\infty ds \delta\left(\sum_{j=1}^n y_j - s\right) $$ $$ = \int_{s=0}^\infty ds \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n}\delta\left(\sum_{j=1}^n y_j - s\right) $$ Make the substitution $y_j = ...



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