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31

Behold the power of symmetry: $$\cos(\pi - x) = - \cos x\quad\text{and} \quad \sin (\pi - x) = \sin x,$$ therefore \begin{align} \int_0^\pi \cos x \log (\sin^2 x + 1) \,dx &= \int_0^\pi \cos (\pi - u)\log (\sin^2(\pi - u) + 1)\,du\\ &= - \int_0^\pi \cos u\log (\sin^2 u + 1)\,du, \end{align} hence the integral evaluates to $0$.


5

\begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta\right)^2\cos ...


5

You may use that, for any continuous function $f$ over $[0,1]$, as $n \to \infty$, we have $$ \frac{1}{n} \sum_{i=0}^n f\left(\frac{i}{n}\right) \longrightarrow \int_0^1f(x)\:dx. $$ Apply it with $$ f(x)=\frac1{1+x^2}, $$ giving, as $n \to \infty$, $$ \sum_{i=1}^n\frac{n}{n^2+i^2}=\frac1n\sum_{i=1}^n\frac1{1+(i/n)^2} \longrightarrow ...


5

In case you're looking for a way that doesn't involve splitting into partial fractions, you can use the substitution $\tan t=x$ so that $\sec^2t\,\mathrm{d}t=\mathrm{d}x$, giving $$\int\frac{2}{x^2(x^2+1)^2}\,\mathrm{d}x=2\int\frac{\sec^2t}{\tan^2t(\tan^2t+1)^2}\,\mathrm{d}t=2\int\frac{\cos^4t}{\sin^2t}\,\mathrm{d}t=2\int(\csc^2t-2+\sin^2t)\,\mathrm{d}t$$ ...


5

Your intuition (expressed in the comments below the OP) is that the integral in question is analogous to $$\lim_{x \to \infty}(x-x)$$ which is, of course, $0$. But a better way to think about it is to consider $$\lim_{(x,y)\to(\infty,\infty)}(x-y)$$ which informally can be thought of as $\infty - \infty$, but this cannot be simply expressed as $0$. More ...


5

Using the Newton-Leibnitz rule, $$\frac{d}{dx} \int_{f(x)}^{g(x)} h(t) dt=h(g(x))g'(x)-h(f(x))f'(x)$$ This should solve the problem.


4

Hint. Observe that if you set $$ f(x)=\int_{-\infty}^{-x} e^{- \frac{1}{2}y^{2}} dy\quad \text{then} \quad f'(x)=- e^{- \frac{1}{2}x^{2}} $$ and the initial integral takes the form $$ \begin{align} \int_{0}^{+\infty} \int_{-\infty}^{-x} \frac{1}{2 \pi} e^{- \frac{1}{2}(x^{2}+y^{2})} dx dy&=-\frac{1}{2 \pi}\int_{0}^{+\infty}f'(x)\cdot f(x)\:dx\\\\ ...


4

Since both factors are repeated, the partial fractions decomposition of the integrand has the form $$\frac{1}{x^2 (x^2 + 1)^2} = \frac{A}{x^2} + \frac{B}{x} + \frac{C x + D}{(x^2 + 1)^2} + \frac{E x + F}{x^2 + 1} .$$ The resulting linear system is somewhat complicated (six equations in six variables), but we can streamline solving it by noticing that the ...


3

Hint For the second: $$\int \frac{x+1}{x^2+x+1}dx$$ Rewrite as $$\int\bigg( \frac{2x+1}{2(x^2+x+1)}+\frac{1}{2(x^2+x+1)} \bigg)dx$$ $$=\underbrace{\frac 1 2 \int\frac{2x+1}{x^2+x+1}dx}_{:=I}+\underbrace{\frac 1 2 \int \frac{1}{x^2+x+1}dx}_{:=J}$$ For $I$ substitute $t=x^2+x+1$ and $dt=(2x+1)dx$ For $J$ complete the square : $\frac 1 2\displaystyle\int ...


3

Rewrite the equation as follows : $$1-f(x+y)=(1-f(x))(1-f(y))$$ Now denote $1-f(x)=g(x)$ so the equation is now : $$g(x+y)=g(x)g(y)$$ Now it follows that $g(x) \geq 0$ because : $$g(x)=g\left(\frac{x}{2} \right )^2 \geq 0$$ If for some $a$ we have $g(a)=0$ then : $$g(x)=g(a)g(x-a)=0$$ for every $x$ so $f(x)=1$ which contradicts the fact that ...


3

$$\int\cos x\ln(1+\sin^2x)\ dx$$ $$=\ln(1+\sin^2x)\int\cos x\ dx-\int\left(\dfrac{d\ \ln(1+\sin^2x)}{dx}\int\cos x\ dx\right)dx$$ $$=\sin x\cdot\ln(1+\sin^2x)-\int\dfrac{2\sin^2x\cos x}{1+\sin^2x}dx$$ $$\int\dfrac{2\sin^2x\cos x}{1+\sin^2x}dx=\int\dfrac{2(1+\sin^2x-1)\cos x}{1+\sin^2x}dx=2\int\cos x\ dx-2\int\dfrac{\cos x}{1+\sin^2x}dx$$ Set $\sin x=u$ ...


3

So, if $\sin x = \cos (\frac{\pi}{2}-x)$, then $$\int_{x=0}^{\pi/2} \sqrt{1 - \sin x} \, dx = \int_{x=0}^{\pi/2} \sqrt{1 - \cos (\tfrac{\pi}{2} - x)} \, dx = \int_{u=0}^{\pi/2} \sqrt{1 - \cos u} \, du,$$ the last equality due to the substitution $u = \pi/2 - x$, $du = - dx$. Now use the same method of solution, namely $\sqrt{1 - \cos u} = \sqrt{2} \sin ...


3

We assume $p^2<1$. First observe that $$ 2I^{\prime} (p)=\!\!\int^{\pi}_{0}\!\! \frac{2\sin x \sin (nx)}{1+p^2-2p \cos x}dx =\!\!\int^{\pi}_{0}\!\! \frac{\cos((n-1)x)}{1+p^2-2p \cos x}dx-\!\!\int^{\pi}_{0}\!\! \frac{\cos((n+1)x)}{1+p^2-2p \cos x}dx. \tag1 $$ Then, by the change of variable $t=\tan (x/2)$, one gets $$ \int^{\pi}_{0} \frac{1}{1+p^2-2p ...


3

For $p^{2} <1$, $$\frac{\sin x}{1+p^{2}-2p \cos (x)} = \sum_{k=1}^{\infty} p^{k-1} \sin(kx). $$ This can be derived by evaluating the geometric series $\sum_{k=0}^{\infty}(p e^{ix})^{k} $. So assuming $n$ is a positive integer, we have $$ \begin{align} I'(p) &= \int_{0}^{\pi} \sin (nx) \sum_{k=1}^{\infty} p^{k-1} \sin(kx) \, dx \\ &= ...


2

Let $h=2n$, and $f$ be convex on $[0,\pi]$. Then $g(x)=\frac1{2n}\,f\!\left(\frac{x}{2n}\right)$ is convex on $[0,2\pi n]$. $$ \begin{align} \int_0^\pi f(x)\cos(hx)\,\mathrm{d}x &=\int_0^\pi f(x)\cos(2nx)\,\mathrm{d}x\tag{1}\\ &=\int_0^{2\pi n}\frac1{2n}\,f\!\left(\frac{x}{2n}\right)\cos(x)\,\mathrm{d}x\tag{2}\\ &=\int_0^{2\pi ...


2

\begin{align} \frac{1}{2}\sin t\left(1-\cos t\right)\sqrt{\frac{1}{2}-\frac{1}{2}\cos t} &= \frac{1}{2}\sin t\left(1-\cos t\right)\sqrt{\frac{1}{2}}\sqrt{1-\cos t} \\ &= \frac{1}{2\sqrt{2}}\sin t\left(1-\cos t\right)^{3/2}\,dt \end{align} so let $u = 1-\cos t\to du = \sin t \,dt$ which makes the problem easier to solve.


2

This may seem tricky at first but by the fundamental theorem of calculus, $$F(b)-F(a)=\int^b_a f(x)~\mathrm dx$$ and by the Second Fundamental Theorem of calculus $$F(x)=\int_a^xf(t)~dt \Rightarrow F'(x)=f(x).$$ So, in this case, all you are really doing is taking the derivative of the right-hand side. $$y(x)=x\int_2^{x^2}\sin(t^3)~\mathrm dt$$ ...


2

The distribution of $(X,Y)$ is rotationally invariant, so the chance that it lies in the region $\{(x,y): x>0, x+y<0\}$ (shaded area below) is $1/8$.


2

Consider, $$j(a) = \int_{0}^{\infty} e^{-ax^2}\sin^2(x)\mathrm{d}x$$ $$= \int_{0}^{\infty} e^{-ax^2}\frac{1-\cos(2x)}{2}\mathrm{d}x$$ $$= \int_{0}^{\infty} e^{-ax^2}\frac{1-\cos(2x)}{2}\mathrm{d}x$$ $$= \frac{\sqrt{\pi}}{4} a^{-1/2} - \frac{\sqrt{\pi}}{4} e^{-1/a} a^{-1/2}. $$ Now let $J(a)$ be an antiderivative of $j(a)$. $$ J(a) = ...


2

We can use Parseval's theorem to solve the integral. The rectangle function is, $$\Pi(x) = \begin{cases} 1 \quad |x|<1/2 \\ 0 \quad \text{otherwise} \end{cases}$$ The Fourier transform of the rectangle function is the $\mathrm{sinc}$ function. $$ \hat{\Pi}(k) = \mathrm{sinc}(k) = \frac{1}{\sqrt{2\pi}}\frac{\sin(k)}{k}.$$ Parseval's theorem tells ...


2

We have: $$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$ hence: $$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$


2

I will consider the integral without the limit process. $$I(a,b)=\int\limits_0^\infty e^{-a x^2}\cos(b x) dx=\sum_{n=0}^{\infty}\frac{(-1)^n.b^{2n}}{(2n)!}\int_0^{\infty}x^{2n}e^{-ax^2}dx$$ by expanding the cosine. The integrals in the above sum are the familiar Gaussian Integrals defined by $$I_m=\int_0^{\infty}x^me^{-ax^2}dx$$ for non negative integral m. ...


2

HINT: the second integral is given by $$\int\frac{x+1}{x^2+x+1}dx$$ you must write the denominator as $$(x+1/2)^2+3/4)$$


2

Here is one approach to the second integral: $$\int \frac{x + 1}{ x^2 + x + 1} dx = \frac12 \left(\int \frac{2x + 1}{ x^2 + x + 1} dx + \int \frac{1}{x^2 + x + 1} dx\right)$$ The first integral is identical to the one you integrated. The second is $\arctan$, which can be seen after you complete the square. $$\int \frac{1}{x^2 + x + 1} dx = \int ...


2

Hint: There's no need to split the integral that way (IMO), we can instead write, $$\begin{align} \int \frac{3x+2}{x^2+x+1}\,\mathrm dx&=\frac{3}{2}\int\frac{2x+\tfrac{4}{3}}{x^2+x+1}\,\mathrm dx\\ &=\frac32\int\dfrac{2x+1+\tfrac13}{x^2+x+1}\,\mathrm dx\\ &=\dfrac32\left(\int\dfrac{2x+1}{x^2+x+1}\,\mathrm dx+\int\dfrac{{\small ...


1

Let $3x+2=A\dfrac{d(x^2+x+1)}{dx}+B$ $\iff3x+2=A(2x+1)+B=2Ax+A+B$ $\implies2A=3,A+B=2$ $$\implies\int\dfrac{3x+2}{x^2+x+1}dx=A\cdot\dfrac{d(x^2+x+1)}{x^2+x+1}+B\dfrac{dx}{x^2+x+1}$$ $$\int\dfrac{dx}{x^2+x+1}=\int\dfrac4{(2x+1)^2+(\sqrt3)^2}dx$$ Set $2x+1=\sqrt3\tan y$


1

The integral can be written as $$ \frac{1}{\sigma^2}\int_b^\infty dx\ x\ e^{-x^2/2}\int_0^\infty dy\ y e^{-y^2 \left(\frac{1}{2}+\frac{1}{2\sigma^2}\right)}I_0(xy)\ . $$ Now we can use the following formula 6.633.4 of Gradshteyn-Ryzhik $$ \int_0^\infty dy\ y\ e^{-\alpha y^2}I_\nu (\beta y)J_\nu(\gamma ...


1

Set $\displaystyle\,\, \cos \theta=\frac{1}{2x^2+1}.$ Then the intagral transforms into $$\frac12\int_0^{\large\frac{\pi}{3}} \sin^{-1}\left(\frac{1-\cos\theta}{2\cos\theta}\right)d\theta.$$ This type of integral is called a Coxeter integral. The user @Sangchul Lee (formerly, sos440) managed to obtain a closed-form formula for a certain family of these ...


1

$\displaystyle J=\int_0^{\tfrac{1}{\sqrt{2}}}\dfrac{\arcsin (x^2)}{\sqrt{1+x^2} (1+2x^2)}dx$ Perform change of variable $y=x^2$, one obtains: $\displaystyle J=\int_0^{\tfrac{1}{2}}\dfrac{\arcsin x}{2\sqrt{x(1+x)}(1+2x)}dx$ $\displaystyle ...



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