# Tag Info

7

Your first method is not correct. I suspect you are trying to use the following incorrect result: $$\int\frac{1}{f(x)}dx = \log f(x)+C.$$ This is not correct in general. The correct form would be $$\int\frac{f'(x)}{f(x)}dx = \log f(x)+C,$$ which you cannot use in your example because the denominator of your integrand is not a linear function of $x$.

6

Hint: For the integral $$\int \dfrac{\tan^2(x)}{x-\tan(x)}\, dx$$ try the substitution $u = x-\tan(x)$, $du = 1-\frac{1}{\cos(x)^2} \, dx = -\tan^2(x)\, dx$ As requested here is the elaboration: With the substitution $u = x-\tan(x)$ we get \begin{align*}du &= ...

4

Consider the integral \begin{align} I = \int x \, \sin^{2}(4x) \, dx. \end{align} Using $2 \sin^{2}(x) = 1 - \cos(2x)$ then \begin{align} 2 I &= \int x (1 - \cos(8x)) \, dx \\ &= \frac{x^{2}}{2} - \int x \, \cos(8x) \, dx. \end{align} Using integration by parts yields \begin{align} \int x \cos(8x) \, dx &= \frac{x \sin(8x)}{8} - \int \sin(8x) \, ...

3

Here's one way to go about it: consider that $1-2\sin^2x=\cos 2x\implies \sin^2x=\frac12(1-\cos2x)\implies \sin^24x=\frac12(1-\cos8x)$. Now perform an integration by parts, using $u=x,dv=\sin^24x=\frac12(1-\cos8x)\:dx$ $du=dx,v=\frac12(x-\frac18\sin8x)$ then, $$\int x\sin^2(4x)\;dx=\int u\:dv=uv-\int ... 3 Using \cos 2A = 1 - 2\sin^2 A we get$$\int x\sin^2 4x\,\mathbb d x = \int \tfrac12 x(1-\cos 8x)\, \mathbb d x$$And$$\int x\cos 8x \, \mathbb d x = \tfrac18x\sin 8x - \tfrac18 \int \sin 8x \, \mathbb d x$$So$$\int x\sin^2 4x\,\mathbb d x = - \tfrac1{16}x\sin 8x +\int \tfrac12 x + \tfrac1{16} \int \sin 8x \, \mathbb d x$$3 The second answer is correct. The mistake in the first method is that your computation of the integral (after substitution) is wrong. It holds: \int\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}u=\frac{2}{4} \sqrt{1+u} +C_2=\frac{1}{2}\sqrt{1+2x^2}+C_2 3 Another one just for fun ;-) Set x=\frac{1}{\sqrt 2}\sinh(u) then,$$\int^t\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt 2 t)}\frac{\cosh(x)\sinh(u)}{\sqrt{1+\sinh^2(u)}}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 t)}\frac{\cosh(u)\sinh(u)}{\cosh u}du=\frac{1}{ 2}\int^{\text{argsinh}(\sqrt2 ...

3

From Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik equation $3.631\ (9)$ we have $$\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}$$ Proof Integrating $(1+z)^p z^q$, for $p,q\ge0$, in the $z=u+iv$ plane around the contour ...

3


2

We have $$f(t) \leq \max_{t\in [x-x\varepsilon,x]} f(t) = f(x)$$ and conclude $$\int_{x-\varepsilon x}^{x} f(t)\,dt \leq \int_{x-\varepsilon x}^{x} f(x)\, dt = f(x) [x-(x-\varepsilon x)].$$

2

Why not directly? Since $$\int\frac1{\sqrt x}dx=\int x^{-1/2}dx=\frac{x^{1/2}}{1/2}+C=2\sqrt x+C$$ we get that for any differentiable (and positive) function $\;f\;$: $$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$ In our case, $$f(x)=1+2x^2\;,\;\;f'(x)=4x\implies\int\frac x{\sqrt{1+2x^2}}dx=\frac14\int\frac{(1+2x^2)'}{\sqrt{1+2x^2}}dx=$$ ...

2

In method 1, you have a mistake. If $$u=2x^2$$ $$x=\frac{\sqrt{u}}{\sqrt{2}}$$ $$dx=\frac{1}{2 \sqrt{2} \sqrt{u}}$$ and $$\int \frac{x}{\sqrt{1 + 2x^2}} dx=\frac{1}{4} \int \frac{du}{\sqrt{u+1}}=\frac{\sqrt{u+1}}{2}$$

2

In first you have: $$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$ It should be: $$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}du=\frac{1}{4}\cdot 2 \sqrt{1+u}+C$$

1

Assuming that $\varepsilon$ and $x$ are both non-negative, let $g$ be the constant function whose value is $f(x)$ everywhere. Then because $f$ is non-decreasing we have $g(t)\le f(t)$ for $t\le x$. Thus, in particular $$\int_{x-\varepsilon x}^x f(t)\,dt \le \int_{x-\varepsilon x}^x g(t)\,dt = \int_{x-\varepsilon x}^x f(x)\,dt = \varepsilon x f(x)$$

1

$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$ \begin{align} y & = x\tan\theta \\ dy & = x\sec^2\theta\,d\theta \\ y^2+x^2 & = x^2(\tan^2\theta+1) = x^2\sec^2\theta \\ y^2-x^2 & = x^2(\tan^2\theta-1) \\ & \phantom{=}\text{etc.} \end{align} Usually with $\displaystyle\int \Big(\cdots\cdots\text{something involving ... 1 Use $$\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C.$$ Note that $$\int\frac{2}{x+2}dx=2\int\frac{(x+2)'}{x+2}dx$$ $$\int\frac{1-2x}{x^2+1}dx=-\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx=-\int\frac{(x^2+1)'}{x^2+1}dx+\int\frac{1}{x^2+1}dx.$$ Here, set$x=\tan\theta$for $$\int\frac{1}{x^2+1}dx.$$ 1 Using Maple I am obtaining$0.4848560378$. Do you agree? 1 To get the total cross section, you also have to integrate over the azimuthal angle. Since there is no azimuthal dependence, that will just result in a factor of $$2\pi$$ to the total cross section, for a final result of $$\sigma = \frac{a^2}{2}\cdot2\pi = \pi a^2$$ This is also the cross-sectional area of the scattering potential. 1 The moment you do an integration by parts, you should add a constant$c_1$to the equation, but you repeated the integration by parts so you would get another constant$c_2$which gets absorbed by the earlier$c_1$. And when you finally solve for the required integral, you'll have the constants piled up on one side which then equal some new constant$c\$. But ...

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