# Tag Info

0

Notice that for $x\to k\pi/2$ that: $$\arctan\left(d+e\tan(x/2)\right)\sim\arctan(\tan(x))=\pi/2$$

1

While, $F(x) = \frac{2 \arctan\left(\frac{a+b \tan \left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}}$ may not be defined at $x = \pi, \lim_\limits{x\to\pi} F(x)$ is defined and we can use that to evaluate the function at the upper bound.

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Well $$\int_{\pi}^{\infty}\frac{\sin^2x}{x^2}dx\leq\int_{\pi}^{\infty}\frac{dx}{x^2}$$ converges, but $$\int_{\pi}^{\infty}\frac{|\sin x|}{x}\sqrt{1+\frac{\cos^2x}{x^2}-\frac{2\sin x}{x^3}+\frac{\sin^2x}{x^4}} \ dx\geq\int_{\pi}^{\infty}\frac{|\sin x|}{x}dx$$ diverges.

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No. One the complex plane, only analytic functions have derivatives or anti-derivatives. And neither $|z|$ nor $|z|^2$ is analytic anywhere. These functions are real-valued everywhere, and the only complex-analytic functions whose images are not 2-dimensional are constants. Integration in $\Bbb C$ is along curves. So to have an integral, you must define the ...

1

An idea for a solution just came to my mind, which I am going to post here hoping it might be useful for somebody in the future. Unfortunately, the condition doesn't work in my case, so that I might go with the Fourier series solution. But for $a\geq b$, we can write $$\frac{1}{a \sin x +b}=\frac{1}{a}\frac{1}{\sin x +\frac{b}{a}}=\frac{1}{a}\frac{1}{\sin x +... 1 Assume b>a such that b+a\sin(x)\ne0. Then, per the Fourier series$$\frac{\sqrt{b^2-a^2}}{b+a\sin(x)}=1+2\sum_{n=1}^{\infty}\left(\frac{\sqrt{b^2-a^2}-b}{a}\right)^n[\sin\frac{n\pi}2\sin{(nx )}+ \cos\frac{n\pi}2\cos{(nx )} ]$$1 Consider$$f(x)=\frac{1}{\sin (x)+k}$$If k\ll 1, you could use expansions and write$$f(x)=\sum_{n=0}^\infty (-1)^n \csc^{n+1} (x)\,k^n$$If k\gg 1, you could use expansions and write$$f(x)=\sum_{n=0}^\infty (-1)^n \sin^{n}(x) \,k^{-(n+1)} $$1  f  is continuous at  I= [0,+\infty) , so it has an antiderivative at  I  given by$$(\forall x\ge 0)\;\;F_1(x)=\cos(x)+C_1$$By the same, it has an antiderivative at (-\infty,0)  given by$$(\forall x<0)\;\; F_2(x)=x-\frac{x^3}{3}+C_2$$F  is continuous at  0 , if C_2=C_1+1.  F(\frac{\pi}{2})=0  if  C_1=0 . So,$$F(-1)+F(\pi)=-1+\frac ...

1

Let $x= \sin^2 t$ so that $dx = 2\sin t \cos t \; dt.$ Then $$\int \frac{1}{\sqrt{x}} \; dx = \int \frac{1}{\sin t} 2\sin t \cos t \; dt = 2\int \cos t \; dt$$ $$=2\sin t + C = 2\sqrt{x} + C.$$

3

for $x>0 \$ with integration by parts $$u=\frac{1}{x} , \ \ u'=\frac{-1}{x^2} \\ v=\frac{2}{3}x^\frac{3}{2} \ \ \ \ v'=\sqrt{x}$$ \begin{align} \int \frac{1}{\sqrt{x}}dx =\int \frac{\sqrt{x}}{x} dx & = \frac{2}{3}x^\frac{3}{2}.\frac{1}{x}-\int \frac{-1}{x^2}.\frac{2}{3}x^\frac{3}{2} dx +c\\ & = \frac{2}{3}\sqrt{x}+ \int \frac{2}{3} \frac{\... 0 Evaluate \int \frac{1}{\sqrt{x}} \; dx without using the power rule Consider the u-substitution formula backwards by treating u as the dummy variable in the integral on the right such that\int \frac{1}{\sqrt{\phi(x)}} \; \phi'(x) \; dx = \int \frac{1}{\sqrt{u}} \; du$$and if you find the antiderivative on the left with respect to x then you have ... 1 Your answer is ok. But it is better to integrate$$ \frac{1}{(t+1)^2} \leq \frac{1}{t+1}$$from 0 to x; namely$$ \int_0^x\frac{1}{(t+1)^2}dt \leq \int_0^x\frac{1}{t+1}dt. $$Then you will get the desired inequality soon instead of concluding C_1=C_2=0. 0 Use x^x=e^{x \log(x)}$$x^x=\sum_{n=0}^\infty \frac 1{n!} \big[x \log(x)\big]^n\int_0^a \big[x \log(x)\big]^n \,dx=- \log^{n+1} (a) \,\,E_{-n}(-(n+1) \log (a))$$Computing the partial sums \sum_{n=0}^p for a=2, this gives$$\left( \begin{array}{cc} p & \sum_{n=0}^p \\ 0 & 2.000000000 \\ 1 & 2.386294361 \\ 2 & 2.707063849 \\ ...

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According to the Dirichlet test for uniform convergence, the improper integral $\int_0^\infty f(x,t) g(x,t) \, dx$ is uniformly convergent for $t \in D$ if $\int_0^x g(y,t) \, dy$ is uniformly bounded for all $x >0$ and $t \in D$ and $f(x,t) \searrow 0$ as $x \to \infty$ monotonically with respect to $x$ and uniformly for $t \in D$. In this case, the ...

1

Write $r = r(\theta) = \sqrt{1 - 2x\cos\theta + x^2}$. Then we easily check that $$\frac{\mathrm{d} r^{\alpha}}{\mathrm{d}\theta} = \alpha r^{\alpha-2} x\sin\theta, \qquad\text{and so},\qquad \int r^{\beta}\sin\theta \, \mathrm{d}\theta = \frac{r^{\beta+2}}{(\beta+2)x}.$$ Now denoting the integral by $f(x)$ and assuming that $x \neq \pm 1$, we have \begin{...

0

Consider the sequence of substitutions: $$u=\cos\theta-x\implies\mathrm du=-\sin\theta\,\mathrm d\theta$$ $$\implies f(x)=-\int_{1-x}^{-1-x}\frac u{(u^2+1-(u+x)^2)^{\frac32}}\,\mathrm du=\int_{-1-x}^{1-x}\frac u{(1-2ux-x^2)^{\frac32}}\,\mathrm du$$ Then $$v=1-2ux-x^2\implies\mathrm dv=-2\,\mathrm du$$ $$\implies f(x)=\frac1{4x}\int_{x^2+2x+1}^{-3x^2+2x+1}\... 0 Actually, \int_{-\infty}^0\delta(t)dt is undefined, because \lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt is not the same for all choices of PDFs \delta_0 we can use as a nascent delta. For example,$$\begin{align}\delta_0(u)&:=e^{-u}1_{[0,\,\infty)}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=0,\,...

1

To get the given form (which is returned by Wolfram Alpha), rewrite $2-\cos t=1+2\sin^2\frac t2$ and then substitute $u=\sin\frac t2,\frac{du}{dt}=\frac12\sqrt{1-u^2}$: $$\int_0^\varphi(2-\cos t)^{-3/2}\,dt=2\int_0^{\sin\varphi/2}\frac1{(1+2u^2)\sqrt{(1+2u^2)(1-u^2)}}\,du$$ This is an elliptic integral of the third kind (argument convention as in Mathematica/...

1

Around $x=0$ $$\frac{x\cos^3(x)}{t^2+x^2}=\frac{x}{t^2}-\left(\frac{1}{t^4}+\frac{3}{2 t^2}\right) x^3+O\left(x^4\right)$$ so no problem at the lower bound. You can also compute the antiderivative writing $$\frac{x}{t^2+x^2}=\frac{x}{(x+it)(x-it)}=\frac{1}{2 (x+i t)}+\frac{1}{2 (x-i t)}$$ and use $$4\cos^3(x)=\cos(3x)+3\cos(x)$$ to face rather simple ...

5

As zkutch has given how to calculate with absolute value here is another approach To avoid confusion its better to do this we have $$\int \frac{\cos x}{\cos x}\sqrt{1+\sin x}dx=\int \frac{\cos x dx}{\sqrt{1-\sin x}}=-2\sqrt{1-\sin x}+C$$ Here in above integral i had put $1-\sin x=t$ which gave answer easily

1

Hint: $|\sin x|=(-1)^n \sin x$ for $\pi n \leqslant x < \pi (n+1)$, where $n=0, \pm 1,\pm2, \cdots$. To keep antiderivative continuous in $x=\pi (n+1)$ you need to solve $$\left((-1)^{n+1}\cos x +C_n\right) \Big|_{x=\pi (n+1)}=\left((-1)^{n+2}\cos x +C_{n+1}\right)\Big|_{x=\pi (n+1)}$$ taking some $C=C_0$.

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And, adding to NicNic8's answer... There's a problem writing $$\int \frac{\mathrm{d}u}{u} = \ln u + C$$ and that problem is that the left-hand side is defined for any $x \neq 0$ and the right-hand side is defined for any $x > 0$. The thing on the left is defined where the thing on the right is not. So, as per normal, we are using the notion of ...

0

The derivative of the natural logarithm is 1 over its argument. That is, if $f(u) = \ln(u)$ then $f'(u) = \ln'(u) = 1/u$. Taking the integral of both sides, we get $\int \frac{du}{u} = \ln(u) + c$ where $c$ is any constant.

0

Let $x_n$ be the successive values of $x$ where $f(x)$ takes an integer value (they can be obtained by solving $f(x)\in\mathbb Z$ and sorting). As $\left\lfloor f(x)\right\rfloor$ is piecewise constant, the antiderivative is piecewise linear, continuous, with a slope equal to $f(x_k)$. If we start integrating at $x_0$, up to some $x_n$, the integral is the ...

1

$1)$ The area under the graph of $\lfloor{x}\rfloor$ is $0+1+2 +\cdots+(x-1)$ so $$\int{\lfloor{x}\rfloor} \, dx =\frac{x(x-1)}{2}$$ $2)$ The graph of $\lfloor{\sin(x)}\rfloor$ alternates between $0$ and $-1$ every $\pi$ values, so every $2\pi$ values the integral will $-1$. So $$\int\big\lfloor{\sin(x)}\big\rfloor \, dx = -\lfloor{\frac{x}{2\pi}}\rfloor-\... 0$$I=\int_{1}^{\infty}\frac{dx}{x^p}=\left .\frac{x^{1-p}}{1-p}\right|_{1}^{\infty}=\frac{1}{p-1}, ~if~ p>1,$$because 0^{1-p}=0 if only p>1, otherwise it is infinite. Hence the integral converges when p>1. 0 @KaviRamaMurthy and @player2326 have answered this question. The comparison test can be employed to solve this question. Additional references: How to see this improper integral diverges? Checking whther the integral \int_1^∞ \frac{1}{x^{\frac{1}{x}+1}} dx convergent 0 There is no problem in this; it's just a matter of choice. HOWEVER, if you resolve the integrand as$$\frac{2x-3}{(x^2-1)(2x+3)}=\frac{Ax+B}{x^2-1}+\frac{C}{2x-3}$$then how would you go about integrating (Ax+B)/(x^2-1)? Unless you are lucky that B=0, (in the above case you get B=1), in all other cases, you will have no choice but again resolving it ... 2 It does not work because the equation is not an exact DE with M_y=N_x. You would need to find an integrating factor. But you can get an easier result by remarking that the terms are all quadratic or contain the derivative, so that you get the form of an homogeneous DE. Thus set y(x)=xu(x) to find that$$ x^2(−4u+1−u^2+4(xu'+u))=0. As that is now a ... 1 First make the substitution x=\pi + y with y\in (-\pi, +\pi). The integral becomes \begin{equation} \int \frac{d y}{3 - \sin(y)-\cos(y)} = \frac{2}{\sqrt{7}}\arctan\left( \frac{4\tan\left(\frac{y}{2}\right) - 1}{\sqrt{7}}\right) + C \end{equation} after integration with the \tan(y/2) substitution. Now replace y with x-\pi and the solution is ... 0 Making the substitution t=\tan(x/2) gives \begin{align}I(x)&=\frac{1}{2}\int\frac{1+t^2}{2+t+t^2}\frac{2}{1+t^2}\,dt=\int\frac{1}{2+t+t^2}\,dt=\frac{4}{7}\int\frac{1}{1+(\frac{2t+1}{\sqrt7})^2}\,dt\\&=\frac{2}{\sqrt7}\arctan(\frac{2t+1}{\sqrt7})=\frac{2}{\sqrt7}\arctan(\frac{2\tan(x/2)+1}{\sqrt7})\end{align} The problem with this is that this ... 0 Take \sin x=2t/(1+t^2), \cos x=(1-t^2)/(1+t^2), tan(x/2)=t \implies 2 \sec^2(x/2) dx=dt. ThenI=\int \frac{dx}{3+\sin x+ \cos x}=\frac{1}{4}\int \frac{dt}{t^2+t+2}=\frac{1}{4}\int \frac{dt}{(t+1/2)^2+7/4}==\frac{1}{2\sqrt{7}} \tan^{-1} \frac{2t+1}{\sqrt{7}}+C.$$where t=\tan(x/2). 12$$\frac{\sin 3x}{\cos 7x\cos 4x} = \frac{\sin (7x-4x)}{\cos 7x\cos 4x}=\frac{\sin 7x\cos 4x - \cos 7x \sin 4x}{\cos 7x \cos 4x} = \tan 7x - \tan 4x$$I think you can take it from here. 3 Your calculation of the derivative is incorrect. Note$$\begin{align} \frac{d}{dx}\left[ x + \sqrt{x^2 + r^2} \right] &= 1 + \frac{1}{2}\left(x^2 + r^2 \right)^{-1/2} \cdot \frac{d}{dx}\left[x^2 + r^2\right] \\ &= 1 + \frac{1}{2 \sqrt{x^2 + r^2}} \cdot 2x \\ &= 1 + \frac{x}{\sqrt{x^2 + r^2}}. \end{align}You are missing the x in the ... 3 HINT What about the substitution x = r\sinh(u)? In such case, one has that \begin{align*} \int\frac{1}{\sqrt{x^{2}+r^{2}}}\mathrm{d}x & = \int\frac{r\cosh(u)}{\sqrt{r^{2}\sinh^{2}(u) + r^{2}}}\mathrm{d}u = \int1\mathrm{d}u = u + c \end{align*} Can you take it from here? EDIT Since the \sinh function is bijective, there is always a solution to \begin{... 4 The substitution u=1+\sqrt{x} works since dx=2(u-1)du and then you can integrate by parts. 1 This should follow directly from the Leibniz integral rule. With\psi(r)=\int_r^\infty \underbrace{e^{-(y-r)} \left( \int_0^y \psi(y-x)\,\mathrm dx\right)}_{f(r,y)} \,\mathrm dy$$the derivative would be$$\begin{align} \frac{\mathrm d\psi}{\mathrm dr}&=-f(r,r)\frac{\mathrm dr}{\mathrm dr}+\int_r^\infty \frac{\partial f(r,y)}{\partial r}\,\mathrm dy\\[...

1

For $b=0$, the integral can be evaluated exactly in terms of Owen's T function (https://en.wikipedia.org/wiki/Owen%27s_T_function), \begin{equation} I = -\frac{\sqrt{\pi}}{\sqrt{2}a} \left[1+4 T\left(ax,\frac{1}{a}\right) \right]. \end{equation} For $b\ne0$, no idea...

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Another solution kindly suggested by @Raymond Manzoni $$I=\int x\sqrt{\cos (x)}\,dx=\int x\sqrt{1-2 \sin ^2\left(\frac{x}{2}\right)}\,dx$$ Let $x=2y$ $$I=4 \int y \sqrt{1-2 \sin ^2(y)}\,dy$$ Integration by parts $$u=y \qquad v'=\sqrt{1-2 \sin ^2(y)}$$ gives $$\frac 14 I=y\,E(y|2)-\int E(y|2)\,dy$$ Using the expansion $$E(y|2)=y-\frac{y^3}{3}-\frac{y^5}{30}-\... 1$$\cot(\sin^{-1}(w/3))=\cot \theta , \sin \theta =w/3$$So$$\cot \theta= \frac{\sqrt{9-w^2}}{w}.$$Eventually both answers are the same. 2 Note$$\cot(\sin^{-1}\frac{w}{3})= \cot(\csc^{-1}\frac{3}{w}) =\sqrt{\csc^2(\csc^{-1}\frac 3w)-1}\\ = \sqrt{(\frac 3w)^2-1}= \frac{\sqrt{9 - w^{2}}}{w} $$2 Let \theta=\sin^{-1}\frac{w}3. Then \sin\theta=\frac{w}3, which means \cot\theta=\frac{\sqrt{9-w^2}}w. Draw a right triangle where w is the opposite side and 3 is the hypotenuse, since \sin\theta=\frac{w}3. You will find by the Pythagorean theorem that the adjacent side is \sqrt{9-w^2}, which gives the above \cot\theta. 2 You wish to simplify:$$-\cot(\arcsin(\frac{w}{3})) - \arcsin(\frac{w}{3}) + C-\frac{\cos(\arcsin(\frac{w}{3})}{\sin(\arcsin(\frac{w}{3})} -\arcsin(\frac{w}{3}) +C$$Note that \cos(\arcsin(x)) = \sqrt{1 - \sin^{2}(\arcsin(x))} = \sqrt{1 -x^{2}} for x in \arcsin's domain:$$-\frac{\sqrt{1 - \frac{w^{2}}{9}}}{\frac{w}{3}} - \arcsin(\frac{w}{3}) + C$$... 0 Another approach: Let$$I=\int \sqrt{\frac{x-1}{x^{5}}}dx$$so \begin{eqnarray*} I&=&\int \frac{\sqrt{x-1}}{x^{5/2}}dx\\ &=&2 \int \frac{\sqrt{u^{2}-1}}{u^4}du, \quad u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx\\ &=&2\int\sin^{2}(s)\cos(s)ds, \quad u=\sec(s), du=\tan(s)\sec(s)ds\\ &=&2\int p^{2}dp, \quad p=\sin(s), dp=\cos(s)ds\\ &... 4 Hint we have$$\int \frac{1}{x^2}\sqrt{1-\frac{1}{x}}$$Now put$$t=1-\frac{1}{x}$$can you end it now? 5 As said, at the price of the "monster" given by @Raffaele in comments, you can compute it. What you could also do (but at the price of another infinite summation, is to use$$\sqrt{\cos(x)}=t \implies x=\cos ^{-1}\left(t^2\right)\implies dx=-\frac{2 t}{\sqrt{1-t^4}}\,dt$$to make$$I=-2\int \frac{t^2}{\sqrt{1-t^4}}\, \cos ^{-1}\left(t^2\right)\,dt$$... 1 Wolfram's calculation is ugly: https://www.wolframalpha.com/input/?i=integrate+x*sqrt%28cos%28x%29%29+dx But to answer your question directly: yes it can be solved. 0 [F(x)-H(x)] from -2/3 to 1, where H'=h and F'=f u can see the desired area in graph-picture 0 You have a mistake in your h(x) and intersection points. Curves intersect at two points. h(x) = (x+1)^3 - 2 (x+1)^2 - (x+1) + 2 = x^3 + x^2 - 2x Please see the graph for area bound by them. You can use single integral of function (f(x)-h(x)) between bounds of x to find area. 0 I think you are supposed to find this area (f(x) is in red, h(x) is in blue): First, we wish to find the 2 intersection points. Note that h(x) = f(x + 1) = x^{3} -x^{2} + 2x. Then, we want to find:$$f(x) = h(x)x^{3}-2x^{2} - x + 2 = x^{3} + x^{2} -2x3x^{2} - x - 2 = 0(3x + 2)(x - 1) = 0x = -\frac{2}{3}, 1 Note that since \$f(x)&...

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