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• 754

### Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$

Since @Vladimir_Reshetnikov gave an approach by transforming the integral into $$\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\, dy$$ I kinda ...
• 754

### $I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$

Let $x=\cos2a$ for notational convenience, i.e. $$I(a) = \int_0^1 \frac{1}{z}\ln\bigg(\frac{1-z\cos2a+ \sqrt{1-2z \cos2a+ z^2}}{2}\bigg) dz$$ and substitute $z-\cos2a =\sin2a\sinh t$ to remove the ...
• 102k
1 vote

• 23.7k
1 vote

### How to evaluate $\int_0^1 \frac{\arctan(x)}{x} \ln^2(1 - x) \, dx$

Here is an option, however, not a closed-form solution. Expand $\frac{\arctan(x)}{x}$ into a series, i.e., $$\frac{\arctan(x)}{x}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{2n+1}.$$ Thus, you need to ...
• 1,569

### Find: $\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos^2(x) \cos(t)}{(b^2 - a^2 \sin^2(x))(1 + \sin^2(t) \cos^2(x))} \ dt \ dx$

Let $$J(t)= \int_0^{\pi/2} \frac{\cos x\tan^{-1}(\tan t\cos x)}{1-\sin^2s\sin^2x}dx$$ along with \begin{align} J’(t)=& \int_0^{\pi/2} \frac{\sec^2s\sec^2t\cos^2x}{(1+\tan^2s\cos^2x) (1+\tan^2t\...
• 102k

• 270k
Accepted

### How to evaluate $\int_0^\infty \tanh\left(\frac{\pi x}{2}\right) \left(\frac{1}{x} - \frac{x}{x^2 + y^2}\right) \, dx$

Let's consider $$I(a,b)=\int_0^\infty \tanh\left(\frac{\pi x}{2}\right) \left(\frac{x}{x^2 + a^2}- \frac{x}{x^2 + b^2}\right) \, dx$$ then the initial integral $\,\displaystyle I_0=I(a=0, b=y)$. ...
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• 26.3k

### Evaluate $\int^{\infty}_{0} \arctan\left({\frac{a^2}{x^2}}\right)\cos(bx)dx$

WLOG assume $a,b>0$. Sketch of a solution. Integration by parts with $dv= \cos(bx)$. The boundary terms vanish. The new integral is $$\frac{2a^2}{b}\int_{0}^{\infty}\frac{x\sin(bx)}{a^4+x^4}\,dx$$...
• 8,047
Accepted

### Evaluate $\int^{\infty}_{0} \arctan\left({\frac{a^2}{x^2}}\right)\cos(bx)dx$

As already observed by J.G., you special integral has the right result $\pi e^{-\vert b\vert}\frac{\sin(\vert b\vert)}{\vert b\vert}$. I suggest to use a change of variable to use the integral that ...
• 9,150
1 vote

### Closed form of the following sum of binomial coefficients.

Let us define the sequence $a(n)$ by $$a(n) = \sum_{\begin{array}{cc} 1\le k\le n-1\\ gcd(k,n)=1 \end{array}}{n\choose k}$$ It is perhaps worth mentioning that the sequence $a(n)$ is ...
• 133k

### Closed form of the following sum of binomial coefficients.

This is sequence A056188 in OEIS (Encyclopedia of Integer Sequences). No closed-form formula is mentionned in the corresponding article. Here is a (log-scale) representation of the sequence (blue ...
• 84.5k

### Show that $\int_{0}^{\frac{\pi}{2}} \log(1 + \tan^4 x) \cos^2 x \, dx = \frac{\pi \log(6 + 4\sqrt{2})}{4} - \frac{\pi}{2}$

\begin{aligned} \int_0^{\pi/2} \ln(1+\tan^4 x) \cos^2 x \, dx &= \left(\int_0^{\pi/2} \frac{\ln(1+\tan^4 x)}{(1+\tan^2 x)^2} \, d\tan x \Big|_{0}^{\pi/2}\right) \Bigg|_{x \to \tan x} \\ &= \...
• 5,196
1 vote

### How to evaluate $\int_0^1 \frac{x}{1+x^2} \ln ^2\left(\frac{x}{1-x}\right) d x$

Incomplete solution. Let $n\geq 2$, an integer. \begin{align}J_n&=\int_0^1 \frac{x\ln^n\left(\frac{x}{1-x}\right)}{1+x^2}dx\\ &\overset{u=\frac{1-x}{1+x}}=\int_0^1 \frac{\ln^n\left(\frac{1-y}{...
• 14.2k
1 vote
Accepted

• 3,624
Accepted

• 3,510

### Can we further simplify the closed solution to the Goat Problem?

Because there isimplied some complex trigonometric function algebra, use Mathematica ...
• 3,666
1 vote

### How to evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\cos(x)}{(1 + \sqrt{\sin(2x)})^n} \,dx$

\begin{align} &\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1 + \sqrt{\sin2x})^n} \,dx \\ = &\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos x + \sin x}{(1 + \sqrt{\sin2x})^n} \,dx =\int_{\frac\...
• 102k
1 vote

• 23.7k
1 vote

More generally, the integral $$\int_{0}^{\pi/2} \sin(ax) \operatorname{artanh}(\sin 2x) \, \mathrm dx$$ can be expressed in terms of the digamma function. Assume that $0 < a< 2$. For $|r| &... 6 votes ### How to evaluate$\int_{0}^{\frac{\pi}{4}} \frac{\ln(\sin x) \ln(\cos x)}{\tan(2x)} \sin(2x) \, dx Alternatively \begin{align} &\int_{0}^{\frac{\pi}{4}} \frac{\ln(\sin x) \ln(\cos x)}{\tan2x} \sin2x\, dx\\ = &\ \frac12\int_{0}^{\frac{\pi}{4}} {\ln(\sin x) \ln(\cos x)}d\left(\sin 2x\right)\\ ... • 102k 4 votes ### Closed form for \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $The integral$I_n= \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx has the closed form I_n=\frac{(n-1)!!}{(n+2)!!}\bigg(\ln2 + \frac{(-1)^n}{n+2}+\sum_{k=1}^{n} \frac{(-1)^{k}}{k}\bigg) \bigg( \sin^2\frac{n \... • 102k 2 votes ### How to find closed form \int^1_0\frac{\arctan(x)}{1+3x^2}dx More generally \begin{align}\int_0^{1}& \frac{\tan^{-1} x}{1+a^{2}x^2}{dx} =\int_0^1 \int_0^1\frac x{(a^{2}x^2+1)(y^2x^2+1)}dy\ dx\\ &\>\>\>\>\> = \frac12\int_0^1 \frac{\ln \... • 102k 1 vote Accepted ### How to find closed form \int^1_0\frac{\arctan(x)}{1+3x^2}dx HintI=\int\frac{\log(1+y^2)}{3-y}\,dy=\int\frac{\log(y+i)}{3-y}\,dy+\int\frac{\log(y-i)}{3-y}\,dy$$Use$$\int \frac{\log(y+a)}{b-y}\,dy=-\log (a+y)\, \log \left(\frac{b-y}{a+b}\right)-\text{Li}_2\... • 270k 3 votes ### How to evaluate\int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx $(NOTE: Too long for comment) (NOTE: I have utilized "Possible Closed forms" from WA in$J_1,J_3$) $$I = \int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\... • 3,578 3 votes ### How to evaluate \int_{0}^{1} \int_{0}^{1} \frac{4y \, da \, dy}{(y^2 + ay + 1)(y^2 - ay + 1)}? Ramanujan found that$$\int_{0}^{\pi/2} \ln \left( \frac{2 + \sin x}{2 - \sin x} \right) \, dx \, = \, \color{blue}{\displaystyle\sum_{n=0}^\infty \displaystyle\frac{1}{\displaystyle\binom{2n}{n}\left(... • 18.1k 3 votes ### How to evaluate$\int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx $Partial answer $$J_2 = \int\log\left(e^{-x^2}\right) \log\left(\frac{1 + \cos(4x)}{\cos(x)}\right) \, dx= - \int x^2 \log\left(\frac{1 + \cos(4x)}{\cos(x)}\right) \, dx$$ can be computed using a ... • 270k 0 votes ### A closed form of$\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$First, substituting$x \to x^\phi$: $$\int_{0}^{\infty} \frac{x \arctan(x^\phi)}{(x^{\phi^2} + 1)^2} \, dx$$ $$=\phi \int_{0}^{\infty} \frac{x^\phi \cdot \arctan(x^\phi)}{(x^{\phi + 1} + 1)^2} \, dx$$... • 5,196 1 vote ### Prove that$\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} Similar to user153012’s solution \begin{align*} \int_{0}^{1} x^n \sin(\pi x) \, x^x \, (1 - x)^{1 - x} \, dx &= b_{n+2} \pi e, \\ b_1 &= \frac{1}{2}, \\ b_n &= \frac{1}{n} \left( \frac{1}{... • 5,196 4 votes ### Closed form for \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $Noticing that $$I_n=\int_0^1 x^n \sqrt{1-x^2} \ln x d x=I^{\prime}(n)$$ where $$I(a)=\int_0^1 x^a \sqrt{1-x^2} d x$$ Via the substitution$x=\sin \theta\$, we have \begin{aligned} I(a) & =\... • 23.7k 1 vote ### Closed form for \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx  Making the substitution x=\cos\theta you get\int_0^{\frac{\pi}{2}} \cos^n(\theta)\sin^2\theta\ln(\cos\theta)\,\mathrm d\theta$$Furthermore, we can substitute the Fourier series of \ln(\cos\... • 3,624 4 votes ### Closed form for \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx  This is too long for a comment. Gradshteyn and Rhyzik's table of integrals provides the following expansions:$$\int_0^1 x^{2n}\sqrt{1-x^2}\ln x\,dx=\frac{(2n-1)!!}{(2n+2)!!}\cdot\frac{\pi}{2}\left(\...
• 17.6k

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