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2

Logs and inverse trig functions, when part of a more complicated integral, tend to cause problems. By differentiating them, you get 'simpler' more algebraic expressions, because $$\frac{d}{dx}\ln x=\frac{1}{x}$$ and $$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$$ (the other inverse trig functions have similar derivatives with square roots in them). If you ...


0

Put $$\frac{1}{x-1}=t$$ $\implies$ $$ \frac{dx}{(x-1)^2}=-dt$$ So $$-I=\int \frac{t \,dt}{\sqrt{7t^2+8t+1}}$$ Put $$ 7t^2+8t+1=z^2$$ $\implies$ $$(7t+4)dt=zdz$$ So $$ -7I=\int \frac{7t+4-4 \,dt}{\sqrt{7t^2+8t+1}}=\int \frac{7t+4 \,dt}{\sqrt{7t^2+8t+1}}-\int \frac{4 \,dt}{\sqrt{7t^2+8t+1}}$$$$$$ So $$-7I=z-\int \frac{4 ...


2

Let us consider two integrals $$I_1=\int \frac{\mathrm{d}x}{f(x)+1}$$ $$I_2=\int \frac{\mathrm{d}y}{g(y)+c}$$ If it is possible to make a change of variable such that $$g(y)+c=f(x)+1$$ that is to say $$y=g^{-1}(f(x)+1-c)$$ the denominators of the integrands will become the same. However, we shall have the problem of $\frac {dy}{dx}$; if this is a constant, ...


0

You're going in the wrong direction. Hint: Rewrite the integral as $\frac{1}{9} \displaystyle \int \frac{1}{(1 + \frac{4}{9} x^2)}$ dx Let $u = \frac{2}{3} x$.


0

$$\int\frac{dx}{9+4x^2}=\frac19\int\frac{dx}{1+\left(\frac{2x}3\right)^2}$$ Use Trigonometric substitution, $$\frac{2x}3=\tan\theta$$


0

Let $u=2x+1$. Then $\text du=2$ $$\int \sec(2x+1) \ \text dx=\int \frac{\sec(u)}{2} \ \text du$$ $$=\frac 12\int \sec(u) \, \text du$$ $$=\frac 12\ln|\sec(u)+\tan(u)|+\text C$$ Reverse the substitution $$\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C$$ $$\color{green}{\int \sec(2x+1) \, \text dx =\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C}$$


0

$$\int\sec({2x+1})\,dx=\frac{1}{2}\int\sec({2x+1})\,d(2x+1)$$ Can you take it from here?


1

Set $2x+1=u\implies 2dx=du$ $$\int\sec(2x+1)\ dx=\frac12\int\sec u\ du=\frac{\ln|\sec u+\tan u|}2+K$$ Replace back $u$ with $2x+1$


0

You neglected the chain rule: $$ \frac{d}{dx} \ln|\sec(2x+1)+\tan(2x+1)| = \sec(2x+1)\cdot 2. $$


-1

$$\int\frac{x^2 + 1}{x^3 + 3x + 1} dx=\dfrac{1}{3}\int\frac{3(x^2 + 1)}{x^3 + 3x + 1} dx=\\ \dfrac{1}{3}\int\frac{1}{x^3 + 3x + 1} d(x^3 + 3x + 1)=\dfrac{1}{3}\ln\left|x^3 + 3x + 1\right|+C$$


0

$f=\Phi^{-1}(g(x))$ for any function $g$ with a closed form taking values between 0 and 1 will work. Or use integration by parts to obtain $$b\Phi\bigl(f(b)\bigr)-a\Phi\bigl(f(a)\bigr)-\int_a^b x f'(x) \phi\bigl(f(x)\bigr) dx,$$ which suggests that your problem simplifies to finding values such that the sum of the first two terms has a closed form ...


0

This is the Parallel axis theorem, also known as Steiner's Rule. You can read up on it here: http://en.wikipedia.org/wiki/Parallel_axis_theorem


1

Let $\displaystyle a=\frac{7-t^2}{t^2-1}$ instead of $\displaystyle a=\frac{7-t}{t-1} $ ?? Doing that and simplifying we get, $\displaystyle -\frac{12 t}{\left(t^2-1\right)^2}$ and integrating, we get $$-2 \left(-\frac{3 t}{7 \left(t^2-7\right)}+\frac{2 \log \left(\sqrt{7}-t\right)}{7 \sqrt{7}}-\frac{2 \log \left(t+\sqrt{7}\right)}{7 \sqrt{7}}\right)$$ ...


1

You can use the identity that $$\int e^{kx}dx=\frac{1}{k}e^{kx}+C$$ for complex constants $k$ and $C$. Thus the last integral can be evaluated as follows:- $$\int_{-\infty}^\infty e^{-a|\gamma|} e^{i\gamma(x-vt)} d\gamma = \int_{-\infty}^0 e^{a\gamma} e^{i\gamma(x-vt)} d\gamma+\int_{0}^\infty e^{-a\gamma} e^{i\gamma(x-vt)} d\gamma\\=\int_{-\infty}^0 ...


1

Yes, your work with integration by parts shows that for even powers $t^{2n}$, there is no elementary antiderivative for $t^{2n} e^{-ct^2}$ because there is no elementary antiderivative for $e^{-ct^2}$.


1

What you wrote, integrating by parts, is kind of correct. The integral of the Gaussian is impossible to express in terms of simple functions, but is well known, it is called error function. It has the defining property: $$ \mathrm{erf}\bigg(\frac{x}{\sqrt{2}}\bigg) =\sqrt{\frac{2}{\pi}} \int_0^x e^{-\frac{1}{2}t^2}\,dt. $$ Therefore the antiderivative ...


2

There is no elementary anti-derivative for $e^{-t^2/2}$, but there are several special functions such as the error function $$ \int e^{-t^2/2}\,\mathrm{d}t=\sqrt{\frac\pi2}\,\mathrm{erf}\left(\frac t{\sqrt2}\right)+C $$ Note that $\mathrm{erf}(0)=0$. Your integration by parts is correct. The error function above is defined as $$ ...


1

Using IBP by letting $u=t$, $du=dt$, and $$ dv=te^{-\large\frac12t^2}\ dt\quad\Rightarrow\quad v=\int te^{-\large\frac12t^2}\ dt=-e^{-\large\frac12t^2}, $$ yield $$ \int t^2e^{-\large\frac12t^2}\ dt=-te^{-\large\frac12t^2}+\int e^{-\large\frac12t^2}\ dt=-te^{-\large\frac12t^2}+\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right). $$ The last part is ...


0

Since I spent so much time trying to work this out, I'm going to type the answer even though there are already two perfectly good answers. Hah! Anyway, here's how to do it. Let $$u = b - a \cos x$$ and let $$y^2 = a^2 - b^2 + 2ub = a^2 + b^2 - 2ab \cos x.$$ Notice that then making the change of variables to $y$, we have that $$ \begin{align*} 2ydy = ...


1

Another hint : \begin{align} \frac{3x^{2}-x+2}{x-1}&=\frac{3x^{2}-3x+2x+2}{x-1}\\ &=\frac{3x^{2}-3x}{x-1}+\frac{2x}{x-1}+\frac{2}{x-1}\\ &=\frac{3x(x-1)}{x-1}+2\left(\frac{x-1+1}{x-1}\right)+\frac{2}{x-1}\\ &=3x+2\left(1+\frac{1}{x-1}\right)+\frac{2}{x-1}\\ \end{align}


0

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


2

The integral can be written as $$ \int \frac{\sin x (b-a\cos x)}{(a^2+b^2-2ab \cos x)^{\Large\frac{3}{2}}}\,dx=\int \frac{a\cos x-b}{\sqrt{(a^2+b^2-2ab \cos x)^3}}\,d(\cos x). $$ Let $y=\cos x$, then the integral becomes $$ \int \frac{ay-b}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy=\int \frac{ay}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy-\int \frac{b}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy. ...


1

\begin{align} \int_0^{\large\frac{\pi}{4}} \tan^{2n}x\,dx&=\int_0^{\large\frac{\pi}{4}} \tan^{2n-2}x\tan^2 x\,dx\\ &=\int_0^{\large\frac{\pi}{4}} \tan^{2n-2}x(\sec^2 x-1)\,dx\\ &=\int_0^{\large\frac{\pi}{4}} \tan^{2n-2}x\sec^2 x\,dx+\int_0^{\large\frac{\pi}{4}} \tan^{2n-2}x\,dx\\ &=\int_0^{\large\frac{\pi}{4}} \tan^{2n-2}x\,d(\tan ...


0

Like everyone else has said, it makes life easier. Why it works: ∫f'(u)u' d*x* = ∫f'(u) d*u* (both of them equal to f(u), the first because of the chain rule, the second directly because of the fundamental theorem), so you can use u' d*x* = d*u* as a shorthand to rewrite ∫f'(u)u' d*x* in the formal method of u-substitution.


4

HINT: $$\frac{3x^2-x+2}{x-1}=\frac{(x-1)3x+(x-1)2+4}{x-1}=3x+2+\frac4{x-1}$$


9

Rewrite our integral as $$\int 4x\frac{\sqrt{9-4x^2}}{4x^2}\,dx.$$ Let $9-4x^2=4u^2$. Then $x\,dx=-u\,du$, and we arrive at $$\int \frac{8u^2}{4u^2-9}\,du = \int \left( 2 + \frac{3}{2u - 3} - \frac{3}{2u + 3} \right)\ du.$$ Remark: But the answer to your question about this kind of question is probably trigonometric substitution, $2x=3\sin t$.


1

$$(9-4x^2)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}{n}9^{1/2-n}(-4x^2)^{n}$$ $$\dfrac{(9-4x^2)^{1/2}}{x} = \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n}$$ $$\int \left(\dfrac{9}{x^2}\right)^{1/2-n}(-4)^{n} \;\mathrm{d}x =(-4)^{n}9^{1/2-n}\int {x}^{1-2n}\; \mathrm{d}x = (-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right) + C$$ ...


0

Hint: $~t=\cos x\iff dt=-\sin x,~$ or $~u=a^2+b^2-2ab\cos x\iff du=2ab\sin x$.


0

Let $I_j=\int_0^1 \beta^j e^{2\pi im\beta}$. Now use integration by parts: let $u=\beta^j$ and $v^\prime=e^{2\pi im\beta}$ then $\int uv^\prime = uv - \int u^\prime v$ which will give you an integrand of the form $\beta^{j-1}e^{2\pi im\beta}$. So you can write $I_j$ as a function of $I_{j-1}$. And you know what $I_0$ is, so you can solve it recursively.


2

HINT: $$I_{m+1}=\int_0^{\frac\pi4}\tan^{2m+2}dx=\int_0^{\frac\pi4}\tan^{2m}x(\sec^2x-1)dx$$ $$=\int_0^{\frac\pi4}\tan^{2m}x\sec^2x\ dx-\int_0^{\frac\pi4}\tan^{2m}x\ dx$$ $$=\int_0^1 t^{2m}dt-I_m=\frac{t^{2m+1}}{2m+1}|_0^1-I_m$$ $$\implies I_{m+1}=\frac{1-0}{2m+1}-I_m$$


8

Let $x=\cfrac{3}{2}\sin\theta$, then $dx=\cfrac{3}{2}\cos\theta\,d\theta$. \begin{align} \require{cancel} \int\frac{\sqrt{9-4x^2}}{x}\, dx&=\int\frac{\sqrt{9-4\left(\frac{3}{2}\sin\theta\right)^2}}{\cancel{\frac{3}{2}}\sin\theta}\, \cancel{\frac{3}{2}}\cos\theta\,d\theta\\ &=\int\frac{3\sqrt{1-\sin^2\theta}}{\sin\theta}\, \cos\theta\,d\theta\\ ...


2

Hint: $1-\sin^2t=\cos^2t\iff9-\underbrace{9\sin^2t}_{4\,x^2}=9\cos^2t=(3\cos t)^2$


2

We consider $$I(m,n)=\int_{-1}^{1}x^m P_n(x)dx\tag{1}$$ Since $x^{m}$ belongs to the linear span of $P_0(x),\ldots, P_{m}(x)$, the orthogonality implies that for $m<n$ the integral vanishes. Also, it can be easily inferred from parity that $I(m,n)=0$ if $m+n$ is odd. Therefore we assume henceforth that $m\geq n$ and $m+n$ is even. Now let us use the ...


1

If I understood your question correctly, I would say there are $5$ cases. Assume you have a rational function $\dfrac{p(x)}{q(x)}$, where the degree of $q(x)$ exceeds the degree of $p(x)$. Case $1$: $q(x)$ is a product of distinct linear factors Example: Consider $q(x)=\dfrac{x}{(x+3)(x-1)}$ Case $2$: $q(x)$ is a product of linear factors, where some of ...


1

Another approach is based on the fact that, since $g$ is continuous, the existence of a point at which $|g(x_0)|>0$ implies the existence of a little interval $(x_0-\delta,x_0+\delta)$ where $|g|>0$. Therefore $\int_a^b |g| \geq \int_{x_0-\delta}^{x_0+\delta} |g|>0$.


0

(a) Take a look of Mean Value Theorem for integration. (b) Haow about $f(x)=x$, for $x\in[-1,1]$? (c) You should try $f(x)=\lfloor x\rfloor$, for $x\in[0,1]$.


0

You can check by noting that this integral evaluates to $0$ for $s=-1$ when $0<b<a$...


1

The key thing is how to do partial fraction on $\dfrac{(t-s)^n}{t(t-z)^{n+1}}$ generally by e.g. heaviside cover-up method.


2

The integral can be expressed in terms of a series of Gauss hypergeometric functions. It is doubtfull that it would be possible to go further on this way.


1

You added the tags "calculus" and "integration", so I am assuming that these techniques are not beyond you. Hint: We can use multivariable calculus. Set up a volume (triple) integral with $1$ as the integrand. Using spherical coordinates, what will the bounds of integration be? Of course, remember to include the Jacobian determinant. Alternately, this ...


2

Well, $dx_2 = dv$ and $dx_3 =2udu + 2vdv$, so what is $dx_2 \wedge dx_3$?


0

A line integral of the form $\int_\gamma f(x,y)\,dx$ is simply the line integral of the vector field $\vec F(x,y) = f(x,y)\vec i$ over $\gamma$, so if you understand the geometric interpretation of line integrals of vector fields you have your answer. Similarly for a line integral with respect to $dy$.


1

Ok, thanks for your comments about Stieltjes integral. So it's just like Riemann integral, except that summation points are chewed through some function (in this case floor function). $$\int_{n=1}^{\infty}f(x)d g(x)=\sum_{i=1...\infty}f(c_i \in (x_i,x_{i+1}))(g(x_{i+1})-g(x_i))$$ In my case the right bracket $(g(x_{i+1})-g(x_i))$ will be zero everywhere ...


0

First change the order of integration and evaluate the inner integral which is with respect to $z$ using integration by parts, then advance to evaluate the integral involving the floor function. See related techniques.


1

My first instinct: break $g(z)$ into parts, so $$g(z) = z \sum_{k=1}^{N-1} \int_{t=k}^{k+1} k e^{2 \pi i t z} \, dt = z \sum_{k=1}^{N-1} k \biggl[ \frac{e^{2 \pi i t z}}{2 \pi i z} \biggr]_{t=k}^{k+1}.$$ Then see if you can simplify this sum, and then integrate it with respect to $z$.


2

First, $f(x)sgn(f(x)) = |f(x)|$. Suppose $f$ is never zero on $[a,b]$. Then by continuity, it's either always positive or always negative. Suppose it's always positive, i.e., $f(x)>0\ \forall x$. Then $|f|=f$, and the integral becomes $3\int_a^b f(x)\ dx$. But this is a positive quantity, because $f$ is positive, and so we get a contradiction, because by ...


0

Consider $f(x) = 1/\sqrt{x}$ on $[0,1]$. Then $\int_{[0,1]} f(x)$ does not exist. It is not Riemann integrable in the ordinary sense. However, $\lim_{x \rightarrow 0+} \int_{x}^{1} = 2\cdot 1^{1/2} - 0 = 2$. Also note that the other limit does not exist. So, these three expressions are not equivalent. Note that if it is Riemann integrable in the ordinary ...


2

Your integral: $$\int \frac{1}{1+x} \ \text dx$$ Use the u-substitution $u=1+x$. Then $\text du = \text dx$ $$\int \frac 1u \ \text du$$ The integral of $\frac{1}{u}$ is $\ln|u| + C$ $$\int \frac 1u \ du = \ln|u| + C$$ Reverse the substitution. $$\color{green}{\int \frac{1}{1+x} \ \text dx=\ln\left|1+x \right|+C}$$ Hope I helped!



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