Tag Info

New answers tagged

4

Hint As suggested by Paramanand Singh in a comment, change variable $x=\tan^{-1}(t)$. This makes $$\int \frac{\cos^9 (x)}{\sin^3 (x) + \cos^3 (x)} \,dx=\int\frac{dt}{\left(t^2+1\right)^4 \left(t^3+1\right)}$$ Now, using partial fraction decomposition the new integrand becomes $$\frac{-11 t-5}{16 \left(t^2+1\right)}+\frac{2 t-1}{3 ...


1

What you already made is $$I=\int\frac {x^3}{1+x^5}dx=-\int\frac {dt}{1+t^5}$$ Now $$1+t^5=\prod_{i=1}^5 (t-\alpha_i)$$ where $\alpha_i=-(-1)^{\frac i5}$ which means that, using partial fraction decomposition, $$\frac {1}{1+t^5}=\sum_{i=1}^5 \frac{\beta_i}{t-\alpha_i}$$ $$\int\frac {dt}{1+t^5}=\sum_{i=1}^5 \beta_i \log({t-\alpha_i})$$ which is simple, except ...


0

$$\begin{align} T &= \int_0^a \int_0^\infty J_0(\lambda r) J_1(\lambda a) e^{-\sqrt{\frac{s}{\alpha}+\lambda^2} z} dz dr\\\\ &=\frac {J_1(\lambda a)}{\sqrt{\frac{s}{\alpha}+\lambda^2}}\left(\int_0^a J_0(\lambda r)dr\right)\left(\int_0^\infty e^{-\sqrt{\frac{s}{\alpha}+\lambda^2} z} dz\right)\\\\ &=\frac {J_1(\lambda ...


0

Maple gives the result as $$ {\frac {a \alpha\,{{ J}_{1}\left(\lambda\,a\right)} \left( { { J}_{1}\left(\lambda\,a\right)}\pi \,{{ H}_{0}\left(\lambda \,a\right)}-{{ J}_{0}\left(\lambda\,a\right)}\pi \,{{ H}_{1 }\left(\lambda\,a\right)}+2\,{{ J}_{0}\left(\lambda\,a\right)} \right) }{2({\lambda}^{2}\alpha+s)}} $$ where $H_0$ and $H_1$ are Struve functions. ...


2

We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$ Then, $\displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\infty} \frac{1}{n}\frac{\cos ...


1

The integral of interest $\int_0^{\pi}\frac{dx}{1-\sin x}$ does not converge. Rather, it diverges since $$1-\sin x=-\frac12 (x-\pi/2)^2+O\left((x-\pi/2)^4\right)$$ Even if we interpret the integral in the sense of a Cauchy Principal Value, then we have $$\begin{align} \text{P.V.}\int_0^{\pi}\frac{dx}{1-\sin x}&=\lim_{\epsilon \to ...


1

It appears by symmetry in that sin(x) x = 0 .. pi describes an arc that will define an area that is twice that of the half arc described by sin(x) x = 0 .. pi/2 so the second integral is half the area of the first hence multiplied by two makes them equal areas.


3

Make the substitution $u=\pi-x$; then $$\int_{\pi/2}^\pi\frac{dx}{1-\sin x}=\int_{\pi/2}^0\frac{-du}{1-\sin u}=\int_0^{\pi/2}\frac{du}{1-\sin u}\;.$$


2

Let the integral in question be \begin{align}\tag{1} I_{a} = \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+a t)}{1 + t} \, dt. \end{align} For the case of $a=1$ the following is obtained. By utilizing \begin{align} \int_{0}^{\infty} e^{-s t} \, \ln^{2} t \, dt = \frac{1}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) \end{align} for the change $t \to t+1$ leads to ...


0

This gives us an intuition as to how to proceed, the blue curve is $5 \cos 5x$ and red is the other. The two curves intersect when $$5 \cos 5x = 5 - 5\cos 5x \iff \cos 5x = \frac{1}{2}$$ But $$\cos 5x = \frac{1}{2} \implies x = \frac{\pi}{15}$$ in the range we are concerned in. So the area between the curves is $$\int_0^{\frac{\pi}{15}} 5 \cos 5x - (5 - ...


0

Note that $$5\cos(5x)=5-5\cos(5x)\iff 10\cos(5x)=5\iff \cos(5x)=\frac{1}{2}.$$


0

ok guys thanks a lot The reason i made it dissapear is that I am working with the convolution integral The delta function has a value of 1/dt at t=0 so convolving it... (integral) with limits - infinity to plus infinity of... X(tau)d(t-tau) dtau but my logic was - no matter its location the delta always has the value 1/dt or in this case 1/dtau so I ...


2

One may show that this integral is equivalent to the integral in this problem as follows: First, rewrite $$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$ Then note that $$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x} $$ The integrand is then equal to $$\left [\frac{2}{\pi} \tan^{-1}{\left (\frac{2}{\pi} \left ...


5

Make the branch cut for $\sqrt{z}$ along the positive real axis. The integral over the entire closed curve is $2\pi i$ times the residue at $z=-1$: $$ \begin{align} \oint\frac{\mathrm{d}z}{\sqrt{z}(z^2-1)} &=\oint\frac1{2\sqrt{z}}\left(\frac1{z-1}-\frac1{z+1}\right)\,\mathrm{d}z\\ &=2\pi i\cdot\frac i2\\[9pt] &=-\pi\tag{1} \end{align} $$ The ...


5

If one assumes that $f$ is sufficiently well-behaved (and in particular, differentiable), one can solve this by differentiating under the integral sign (this is sometimes called Feynman's trick). Define $$I(a) := \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx.$$ Differentiating with respect to $a$ (and justifying passing the derivative sign through the ...


5

First notice that $$ \begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4 \sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\log(4)}{2} \int_{0}^{\infty} \Big( 1- x \, \text{arccot}(x) \Big) \, dx + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\pi \log(2)}{4} + ...


4

You can use a deformed keyhole contour by providing a semicircular detour of radius $\epsilon$ along each branch. Then consider $$\oint_C dz \frac{z^{-1/2}}{z^2-1} $$ where $C$ is the deformed keyhole contour. Thus, the contour integral is equal to $$\int_{\epsilon}^{1-\epsilon} dx \frac{x^{-1/2}}{x^2-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} ...


4

If we interpret the integral as the Principal Value integral $$\begin{align} I&=\text{PV}\int_0^{\infty}\frac{dx}{x^{1/2}(x^2-1)}\\\\ & =\lim_{\epsilon\to 0}\left(\int_0^{1-\epsilon}\frac{dx}{x^{1/2}(x^2-1)}+\int_{{1+\epsilon}}^{\infty}\frac{dx}{x^{1/2}(x^2-1)}\right) \end{align}$$ then we may evaluate $I$ using contour integration. We will ...


4

EDIT: Oops. What I wrote was assuming that $f(0)=0$, and that in fact $f(t)\to0$ as $t\to0$ fast enough that $\int_0^1f(t)/t$ converges. In general we should take $\int_\delta^A$, then let $\delta\to0$ and $A\to\infty$. I'll leave the details as an exercise - if you do something very similar to what I did below you actually get $$(f(0)-L)\log(2).$$Sorry. ...


2

$$v=\frac{y}{x}$$. $$y=vx$$. $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$.$$v+x\frac{dv}{dx}=1+v+v^{2}$$ $$x\frac{dv}{dx}=1+v^{2}$$ so $$\frac{dv}{1+v^{2}}=\frac{dx}{x}$$ Hence $\arctan(v)=\ln \vert x \vert +C$


0

If you set $y=\sqrt{x}$ you have: $$\int_0^\infty\frac{2ydy}{y(y^4-1)}=2\int_0^\infty\frac{dy}{y^4-1}$$ Since you are integrating an even function you can write it as: $$\int_{-\infty}^\infty\frac{dy}{y^4-1}=\int_{-\infty}^\infty\frac{dy}{(y-1)(y+1)(y-i)(y+i)}$$ Now you can easily use contour integration to get: ...


7

We have: $$ I = \frac{1}{4}\int_{0}^{+\infty}\left(\frac{1}{2}+\frac{1}{e^t-1}-\frac{1}{t}\right)^2\frac{dt}{\sinh^2(t/2)}$$ where: $$\frac{1}{2}+\frac{1}{e^t-1}-\frac{1}{t} = \sum_{n\geq 1}\frac{2t}{t^2+4\pi^2 n^2}\tag{1}$$ as well as: $$ \frac{1}{\sinh^2(t)}=\frac{1}{t^2}+2\sum_{n\geq 1}\frac{(t^2-n^2\pi^2)}{(t^2+n^2\pi^2)^2}\tag{2}$$ so our integral ...


4

Multiply top an bottom by $\sin(b-a)$ and then write $$\sin(b-a)=\sin[(x-a)-(x-b)]=\sin(x-a)\cos(x-b)-\sin(x-b)\cos(x-a).$$


0

Over the interval $I=\left(0,\frac{\pi}{2}\right)$, $0<\sin x<x$, hence for any $\varepsilon\in I$: $$\int_{\varepsilon}^{\pi/4}\frac{dx}{x\sin(2x)}\geq\int_{\varepsilon}^{\pi/4}\frac{dx}{2x^2}=\frac{1}{2\varepsilon}-\frac{2}{\pi}$$ and your integral is clearly divergent.


0

Note that $$\sin\left(2x\right)\leq 2x $$ in all real range $\left(0,\infty\right) $, then $$\int_{0}^{\pi/4}\frac{1}{x\sin\left(2x\right)}\geq\int_{0}^{\pi/4}\frac{1}{2x^{2}}dx=\infty. $$


0

In basic geometry you take by definition that (for $a,b$ numbers and $C,D$ curves): $$ \int_{aC+bD}\omega\,dx := a\int_{C}\omega\,dx + b\int_{D}\omega\,dx. $$ The motivation for this comes from algebraic topology. So: $$ \oint_{C_1-C_2}F\,dr =\oint_{C_1}F\,dr - \oint_{C_2}F\,dr. $$ You can think of the part $S$ of the cylinder between the two curves as ...


0

multipliying both numerator and denominator by x,you will get $$\int { \frac { x^{ 4 }dx }{ x\left( 1+x^{ 5 } \right) } =\frac { 1 }{ 5 } \int { \frac { dx^{ 5 } }{ x\left( 1+x^{ 5 } \right) } } } \\ $$ now assume $t={ x }^{ 5 }$ then your integral will be $$\int { \frac { dt }{ { t }^{ \frac { 1 }{ 5 } }\left( 1+t \right) } } =\int { t^{ -\frac { 1 ...


0

There is a general closed form for such integrals but it's not elementary or pretty at all. $$\int \frac{x^m}{1+x^n} \, \mathrm{d}x = \frac{x^{m+1} \, _2F_1\left(1, \frac{m+1}{n};\frac{m+n+1}{n}; -x^n\right)}{m+1} + \mathrm{C}.$$ Where $_2F_1\left(a, b;c; x\right)$ is the hypergeometric function. With regards to your specific problem, try partial ...


-2

You can bring it to $\int \frac{dx}{x\sin x}$ integral with some replacements which doesn't converge.


1

You should write like this: $$3\int{\sec^2(3x)tan(3x)dx}$$ put $u=3x;\dfrac{1}{3}du=dx$ $$3\int{\dfrac{\sec^2(u)tan(u)}{3}du}$$ Put $v=sec(u);\dfrac{dv}{sec(u)}=\tan(u)du$ $\int{vdv}=\frac{v^2}{2}+c$, Now replacing back the values,we have the final solution as, $$\dfrac{1}{2}\sec^2(3x)+C$$ Method2: $$3\int{\sec^2(3x)tan(3x)dx}$$ put ...


2

Both answers are correct. In fact you can differentiate your answers with respect to $x$ to see it gives the integrand to check your answer. For your interest, your two answers are equivalent because $$\frac{1}{2} \sec^2(3x) + C_1= \frac{1}{2}(\tan^2(3x) + 1)+C_1 = \frac{1}{2}\tan^2(3x) +C_2$$


0

$$\int 3sec^2(3x)tan(3x)dx$$ use this substitution :$$u=tan(3x)\\du=3(1+tan^(3x)dx=3sec^2(3x)dx\\3sec^2(3x)tan(3x)dx=3sec^2(3x)dx *tan(3x)=\\du *u$$so $$\int 3sec^2(3x)tan(3x)dx=\int udu=\frac{1}{2}u^2=\\\frac{1}{2}tan^2(3x) +const$$


0

Range of $x(1)$ is : $ -2 \le x(1) \le 2$ We split this into two cases: first the case : $-2 \le x(1) \le 0$ . The range of $x(0)$ has then to be : $-1 \le x(0) \le 1+x(1)$ second the case : $0 \le x(1) \le 2$ . The range of $x(0)$ has then to be : $x(1)-1 \le x(0) \le 1$ Then simply integrate $1/4$ over these two intervals in order to obtain the pdf of ...


0

$$\int \frac{q}{\sqrt[3]{k^3-q^3+\frac{3 }{8}q^2}} \, dq$$ First, solve for $r$ the equation : $$k^3-r^3+\frac{3 }{8}r^2=0$$ The three roots (real and complex) are : $r_1$ , $r_2$ and $r_3$. The integral can be written on the form : $$\int \frac{q}{\sqrt[3]{(q-r_1)(q-r_2)(q-r_3)}} \, dq$$ On this form, the integral is known to be related with the Appell's ...


0

You are not convolving the densities. The integral you presented is an application of the following theorem: http://mathworld.wolfram.com/TotalProbabilityTheorem.html In your case the summation step over the events partitioning the sample space is what leads to the integral and the formula for finding the density may be recovered by differentiating the ...


0

Note that for a bounded continuous $h:[0,1]\rightarrow \mathbb{R}$, $$\lim_{n\rightarrow\infty} \frac{1}{n}\log\int_0^1 \exp(-n h(x))\ dx = -\min_{x\in[0,1]} h(x).$$ You can prove this pretty straightforwardly by using continuity of $h$ to bound the integral from below by a value close to the minimum on an open set, and from above by the minimum of $h$. ...


0

In cylindrical coordnates, you have, after some simplification $$z=r^{2}(\cos ^{2}\theta -\sin ^{2}\theta )=r^{2}\cos 2\theta $$ and $$r= a\sin \theta \cos \theta =\frac{\vert a\vert }{2} \sin 2\theta $$ where in the latter equation we have taken the positive root since $- \sin 2\theta = \sin 2(-\theta )$ and so the graph is the same. The graph of the ...


0

Assume that $f_Y$, the pdf of $Y\ge 0$, exists. In this case the expected value is $$E[Y]=\int_{0}^{+\infty}x\ f_Y(x)\ dx$$ Also, assume that the integral above is finite. It is obvious that $x$ can be written as $$x=\int_0^x\ 1\ dy, \ x\ge 0.$$ With this, the expected value is $$E[Y]=\int_{0}^{+\infty}\int_0^x\ 1\ dy\ f_Y(x)\ dx.$$ Since the ...


0

$$u=\tan(\frac{x}{2})\\ \sin x=\frac{2u}{1+u^2} \\ \cos x=\frac{1-u^2}{1+u^2}$$ now , in this case $$u=\tan(\frac{x}{2}) \rightarrow \frac{1}{2}(1+\tan^2(\frac{x}{2}))dx=du \rightarrow dx=\frac{2du}{1+u^2}\\\frac{1}{1+\sin x}dx=\frac{1}{1+\frac{2u}{1+u^2}}\frac{2du}{1+u^2}=\frac{2du}{1+u^2+2u}\\=\frac{2du}{(1+u)^2}$$so $$\int \frac{1}{1+ \sin ...


2

Because the derivative of $\ln(f(x))$ is not $\frac{1}{f(x)}$ for all differentiable function $f$, even if it is true for $f(x)=x-a$ where $a$ is a constant. The derivative of $\ln(f(x))$ is $\frac{f^\prime(x)}{f(x)}$ applying the chain rule.


0

Hint: Try differentiating your second expression on both sides, utlizing FTC.


1

The chain rule is the difference. Note that $\int\frac{du}{u}=\ln|u|$. So, you must have a fraction of the form $u$ on the bottom and the derivative of $u$ on the top. For your second example, $u=1-x^2$, but $du=-2xdx$ is not the numerator.


0

The first integral can be solved by linear substitution $z = x - 2$. The second one is non-linear and does not work. Instead you have $-2 \, \int x/(1-x^2) = \mathrm{ln}(1-x^2) + C$. See https://en.wikipedia.org/wiki/Integration_by_substitution.


2

Hint. $$\int\frac{dx}{1+\cos (\pi/2-x)}=\frac{1}{2}\int\frac{dx}{\cos^2 (\pi/4-x/2)}$$


6

Multiply up and down by $1-\sin{x}$: $$\int dx \frac{1-\sin{x}}{1-\sin^2{x}} = \int dx \left ( \sec^2{x} - \frac{\sin{x}}{\cos^2{x}} \right ) = \tan{x} - \sec{x} + C$$


5

HINT : $$\frac{1}{1+\sin x}=\frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1}{\cos^2 x}-\frac{\sin x}{\cos^2x}$$ For the second, set $t=\cos x$.


0

Using your idea, $\displaystyle\int\frac{\sin x}{1+\sin x}dx=\int\frac{\sin x}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}dx=\int\frac{\sin x-\sin^2 x}{\cos^2 x}dx$ $\displaystyle=\int\frac{\sin x}{\cos^2x}dx-\int\tan^2x \;dx=\int\sec x\tan xdx-\int(\sec^2x-1)dx=\sec x-\tan x+x+C$


2

The point mass at $0$ contributes an impulse of area p to the pdf $f_X(x)$. $$f_X(x) = p\delta(x) + f(x)$$ where f(x) is the part of $f_X(x)$ on (0,a]. When you convolve g with that impulse, you get a copy of g back multiplied by p. That adds on to the convolution of the rest of f with g since convolution is linear. $$h(z) = \int_0^{\infty}[p\delta(x) + ...


3

For the second integral: Note first that this integral is easily done by recognizing that $x=(1+x)-1$, so the integral is really $$\int_0^{\infty} \frac{dx}{(1+x)^5} - \int_0^{\infty} \frac{dx}{(1+x)^6} = \frac14-\frac15=\frac1{20}$$ One may also use the residue theorem. However, one must choose an appropriate contour and integrand. In this case, a ...


0

The integral of above expression if upper limit is 1 and lower limit is 0 when integrating with respect to dx is =4.Without dx this integral is meaningless as $\int$ sign represents summation and you haven,t mentioned with what respect it had been summated over.



Top 50 recent answers are included