New answers tagged

0

$\displaystyle J=\int_0^{\tfrac{1}{\sqrt{2}}}\dfrac{\arcsin (x^2)}{\sqrt{1+x^2} (1+2x^2)}dx$ 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 ...


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 ...


0

Let $h(t) = \sqrt{(1+e)^t}$, and let $$ y = \int_{\cos{x}}^{1} h(t) dt = \int_{1}^{\cos x} -h(t) dt. $$ You want to find $\frac{dy}{dx}$. The problem is the top endpoint, which is $\cos x$. If it were just an $x$, then you'd know exactly how to do it--just apply the Fundamental Theorem of Calculus and you'd be done. So let's make a quick substitution: ...


0

$$\int_0^\pi \cos(x) \log(\sin^2(x)+1)dx$$ The first thing we want to do is to remove the $\log$ from the integral. Performing integration by parts we set $u=\log(\sin^2(x)+1)$ and $dv=\cos(x)dx$ to find $du = \frac{2\sin(x)\cos(x)}{\sin^2(x)+1} dx$ and $v = \sin(x)$. Thus the integral transforms to $$\left.\sin(x)\log(\sin^2(x)+1)\right|_0^\pi - ...


18

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 $0$.


1

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


0

HINT: Substitute $y=sin(x), y' = cos(x)$. Then use $w^2+1 = (1+iw)(1-iw)$ and logarithm laws. The integral over logarithm is easy to find in integral tables.


1

Use the Fundamental Theorem of Calculus: $$\frac{d}{dx} \int_{g(x)}^a f(t) dt = \frac{d}{dx}\big( F(a) - F(g(x)) \big)= -g'(x) f(g(x))$$ So in your case: $$\frac{d}{dx} \int_{\cos x}^1 \sqrt{(1 + e)^t} dt = \sin(x)\sqrt{(1 + e)^{\cos x}}$$


0

Notice, $$\frac {d}{dx}\int_{\cos x}^1\sqrt{(1+e)^{t}}\ dt$$ $$=\frac {d}{dx}\int_{\cos x}^1(1+e)^{t/2}\ dt$$ using Fundamental Theorem of Calculus (F.T.C.), $$=(1+e)^{\frac{1}{2}}\frac {d}{dx}(1)-(1+e)^{\Large \frac{\cos x}{2}}\frac {d}{dx}(\cos x)$$ $$=0-(1+e)^{\Large \frac{\cos x}{2}}(-\sin x)$$ $$=\color{blue}{\sin x(1+e)^{\Large \frac{\cos x}{2}}}$$


0

Here you have Integration Limits that depend on $x$. In this case, the Differentiation of the integral is slightly modified. Use the Leibnitz rule for Integration.


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.


1

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 ...


0

Let first pick $x=y=0$ which yields $$ f(0)=f(0)+f(0)-f(0)f(0). $$ Solving for $f(0)$ yields $f(0)=0$ or $f(0)=1$. Now setting only $y=0$, then $$ f(x)=f(x)+f(0)-f(x)f(0). $$ For $f(0)=0$ this yields $f(x)=f(x)$ and for $f(0)=1$ this yields $f(x)=1$. Since $f'(0)=-1$, then $f(x)=1$ would be impossible, thus $f(0)=0$. Now setting $y=dx$ an infinitesimal ...


0

Do induction on even k and odd k separately. First note that this is true for k=1 and k=2 (This will be the base case). Use integration by parts twice to calculate, $\int_{0}^{\pi}x^{2k}\cos(hx)dx = \frac{2k\pi^{2k-2}}{h^2}+\frac{2k(2k-1)}{h^2}\int_{0}^{\pi}x^{2(k-1)}\cos(hx)dx$ The result then follows by induction on k odd, k even separately.


0

First, we can find $f(0)$ using properties of $f$. Given \begin{equation} f(x+y)=f(x)+f(y)-f(x)f(y), \end{equation} Let $x=y=0$. Then we get $f(0)=0$ or $f(0)=1$. Again, let $x=x$ and $y=0$, then we get $f(0)f(x)=f(0)$. If $f(0)=1$, then $f(x)$ is a constant function, contradicting that $f$ decreases. Thus $f(0)=0$. Second, we can show that $f$ is ...


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 ...


0

As already said in comments, it does not seem that the antiderivative of the integrand exists. Concerning the integral, a CAS found something you will not like very much $$I=\frac{\pi }{4} \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2},\frac{3}{ 2};1\right)$$ where appears the generalized hypergeometric function. ...


1

Intuitive idea: Split the area between the curve and the $x-axis$ into thin vertical rectangular strips of width $\Delta x$. For any $x$ in the domain, as you move from $x$ to $x+\Delta x$, the value of the function will change from $f(x)$ to $f(x + \Delta x)$. Let the height of the strip covering the width $x$ to $x+ \Delta x$ be $f(x)$. Then, the area ...


0

Shade the area between the function and the $x$ axis. The area that is above the x axis counts as positive and the area below the x axis counts as negative. The sum of all the area in these regions on the interval of integration is the integral. This is why $\int_{-1}^1 x^3dx=0$, the LHS is negative and the RHS is positive and they cancel out. You can take ...


0

$\int_{0}^{\pi/2} \sqrt{1-\sin x} \, dx$ $ = 2\int_{0}^{\pi/2} \sqrt{1-\sin x} \, d(x/2)$ $ = 2\int_{0}^{\pi/2} \sqrt{(\sin x/2 -\cos x/2)^2} \, d(x/2) $ $ = 2\int_{0}^{\pi/2} (\sin x/2 -\cos x/2) \, d(x/2)$ that you can further on process taking $ x/2= u $ for substitution.


0

Hint : $\sqrt{1-\sin x}=\sqrt{\left(\sin \frac{x}{2}-\cos\frac{x}{2}\right)^2}=\left|\sin \frac{x}{2}-\cos\frac{x}{2}\right|$


0

Use the expression inside the second identity, $\frac{\pi}{2}-x$, as $2x$ in the first identity. This gives you a new expression for the sine $$ \sin x = 2 \cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right) - 1 = 1 - 2 \sin^2\left(\frac{\pi}{4} - \frac{x}{2}\right). $$ I will leave it to you to find which of these two expressions will be best suited to ...


4

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 ...


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 ...


1

$\displaystyle\frac{1}{x^2(x^2+1)^2}=\frac{x^2+1}{x^2(x^2+1)^2}-\frac{x^2}{x^2(x^2+1)^2}=\frac{1}{x^2(x^2+1)}-\frac{1}{x^2+1)^2}$ $\hspace{.9 in}\displaystyle=\left[\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}\right]-\left[\frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\right]$ $\hspace{.9 ...


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 $$\int\frac{2}{x^2(x^2+1)^2}\,\mathrm{d}x=2\int\frac{\sec^2t}{\tan^2t(\tan^2t+1)^2}\,\mathrm{d}t=2\int\frac{\cos^4t}{\sin^2t}\,\mathrm{d}t=2\int(\csc^2t-2+\sin^2t)\,\mathrm{d}t$$ ...


1

$$ \displaystyle \int_{-\infty}^z e^{\frac{-t^2+2t\alpha\mu}{2\sigma^2\alpha^2}+\frac{\lambda t}{1-\alpha}} dt =e^{A B^2}\int_{-\infty}^z e^{-A (t-B)^2}dt\ , $$ with $A=1/(2\sigma^2\alpha^2)$ and $2AB=\frac{2\alpha \mu}{2\sigma^2\alpha^2}+\frac{\lambda}{1-\alpha}$. Solve for $B$. Next, make the change of variable $\sqrt{A}(t-B)=y$, yielding $$ \frac{e^{A ...


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 ...


1

HINT: $$\int\frac{2}{x^2\left(x^2+1\right)^2}\space\text{d}x=$$ $$2\int\left[\frac{1}{x^2}-\frac{1}{x^2+1}-\frac{1}{(x^2+1)^2}\right]\space\text{d}x=$$ $$2\left[\int\frac{1}{x^2}\space\text{d}x-\int\frac{1}{x^2+1}\space\text{d}x-\int\frac{1}{(x^2+1)^2}\space\text{d}x\right]=$$ $$2\left[-\frac{1}{x}-\arctan(x)-\int\frac{1}{(x^2+1)^2}\space\text{d}x\right]=$$ ...


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 ...


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 ...


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 ...


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 $2x+1=\sqrt3\tan y$


2

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


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 ...


3

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 ...


0

The integral equals zero if you adopt the Cauchy principal value concept.


1

Hint. If you make the change of variable $$ u=\sqrt{x},\quad du=\frac{dx}{2\sqrt{x}}, $$ then $$ \int_1^{\infty}\frac{33e^{-\sqrt{x}}}{\sqrt{x}}\:dx=66\int_1^{\infty}e^{-u}\:du. $$


0

Simple: $\int_{-1}^{1} \frac{dx}{x}$ is not well defined. The area, if it should exist in whatever definition of it you choose, is certainly not given by this expression.


1

You can use Parseval's Theorem: If $f(x)$ and $g(x)$ have respective FT's $$F(k) = \int_{-\infty}^{\infty} dx \, f(x) e^{i k x} $$ $$G(k) = \int_{-\infty}^{\infty} dx \, g(x) e^{i k x} $$ Then $$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$ In your case, $f(x) = \frac{\sin{x}}{x}$ and $g(x) = ...


1

\begin{align} \int_{-\infty}^{\infty}\frac{\sin x}{x^3+x}dx & =\Im\int_{-\infty}^{\infty}\frac{e^{ix}-1}{x^3+x}dx \\ & = \Im\left(\frac{1}{2i} \int_{-\infty}^{\infty}\frac{e^{ix}-1}{x}\left[\frac{1}{x-i}-\frac{1}{x+i}\right]dx\right) \end{align} The integrand on the right decays when $x=z$ is in the upper half plane. So you can close the ...


0

Write $$ \sin^2{x} = \frac{1}{2}(1-\cos{2x}), $$ then $$ I = \int_{-\infty}^{\infty} \frac{1-\cos{2x}}{2x^2} \, dx. $$ Now, you can either consider this as the real part of $(1-e^{2ix})/x^2$, and then do the contour integral as usual with $(1-e^{2iz})/z^2$ around a circle in the upper half-plane, taking care with the simple pole at $z=0$, or you can ...


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

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) = ...


1

The comments on your post already solve the problem, as the integrand looks like something resulting from the quotient rule. But you can also use integration by parts for a more methodical approach. \begin{align} \int \frac{x^2e^x - 2xe^x}{x^4} dx = \int x^{-2}e^xdx - 2\int x^{-3}e^x dx \end{align} Now use integration by parts for the first part ...


1

The limits of integration are from $x=0$ to the next value of $x$ for which $y$ is $0$, as seen in the figure. As $$y=\sin^3(2x)\cos^3(2x)$$ $y=0$ when $\sin(2x)=0$ or $\cos(2x)=0$ Thus $2x=n\pi$ or $2x=\frac{(2n+1)\pi}{2}$ or $x=\frac{n\pi}{2}$ or $x=\frac{(2n+1)\pi}{4}$ The least positive value of $x$ here is $x=\frac{\pi}{4}$, which is the upper ...


0

use the Identity, $\cos^2(2x) = 1 - \sin^2(2x)$ Therefore, we have $\sin^3(2x)[1 - \sin^2(2x)]\cos(2x)$ Now, set $u = \sin(2x)$ and things will go smoothly from here.


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. $$y(x)=x\int_2^{x^2}\sin(t^3)~\mathrm dt$$ ...



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