Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [integration]

For questions about the properties of integrals. Use in conjunction with (indefinite-integral), (definite-integral), (improper-integrals) or another tag(s) that describe the type of integral being considered. This tag often goes along with the (calculus) tag.

Integration is a major part of .

There are two main kinds of integrals:

  • definite integrals (e.g. proper and improper integrals), which often have numerical values
  • indefinite integrals, which group families of functions with the same derivative.

Several techniques to solve integrals have been developed, including integration by parts, substitution, trigonometric substitution, and partial fractions.

Integration can be used to find the area under a graph and find the average of the function. Also, it can be used to compute the volume of certain solids and to find the displacement of a particle.

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Help with $\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x = \frac{1}{2} \sinh^{-1}(1)^2$

I've been trying to do the above integral using elementary methods. So far, I've reduced the integral down to evaluating $\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{\sqrt{1-x^2}} \; \mathrm{d}x$ or $\displaystyle \int_0^1 \frac{\sin^{-1}(x)}{1+x^2}…
Sharky Kesa
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Integral involving $\tan(\ln(x))$

I am trying to solve the following integral: $$I=\int\tan(\ln(x))dx$$ which I initially thought would be easy considering there are many ways to solve both: $$\int\sin(\ln(x))dx$$ $$\int\cos(\ln(x))dx$$ But I appear to be wrong. Upon substitution I…
Henry Lee
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isn't this wikipedia Definite integrals involving rational or irrational expression wrong!?

isn't this wikipedia Definite integrals involving rational or irrational expression wrong!? According to my calculations: $\sin[\frac{(m+1)\pi}{n}]$ while in wiki it is $\sin[\frac{(m+1)}{n}]$ seems $\pi$ missed. here is result of my…
Neo
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Why do integrals "start" at 0?

This is a dumb question and I don't really know how to word it. When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0). I can easily see how the derivative of an integral is given by…
Conyare
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Integrating Logistic Functions

For logistic functions in the form of $\frac{C}{1+Ae^{-bx}}$ where $C,A,b>0$ and $x$ is the independent variable, how does one integrate this function type? since during integration, the denominator is to the power of $(-1)$ and integrating will…
lohboys
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Evaluation of $\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$

I need an approach to analytically evaluating this limit: $$\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$$ Numerically, I see that the answer is $-\pi^3$. Similarly, if I replace $x^2$ with $x^4$, I get $-\pi^5$. I…
James Lambers
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Proving that $\lim\limits_{n\to\infty}n\left( \int_0^1 f(t)\, dt -\frac1n\sum_{k=0}^{n-1}f\left(\frac k n\right) \right)=\frac{f(1)-f(0)}{2}$

Let $f\in C^2([0,1])$. Prove that $$ \lim_{n\to+\infty}n\left( \int_0^1 f(t)\, dt -\frac1n\sum_{k=0}^{n-1}f\Big(\frac k n\Big) \right)=\frac{f(1)-f(0)}{2}. $$ The second term is clearly the Riemann sum of the function $f$; since the function…
Romeo
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Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$

Integrate $$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$ Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$ $x=1\rightarrow u={\pi\over 4}$ $x=0\rightarrow u=0$ $$I={1\over 2}\int_{0}^{{\pi\over…
gymbvghjkgkjkhgfkl
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Evaluate :$\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx$

How to evaluate $$ \int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $$ I know that $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ but after that I have no idea, so please help me. Thanks in…
Siddhant Trivedi
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how to solve double integral of a min function

I came across this integral in a book. I don't understand how the author writes the following expression for it. $$\int_0^T \int_0^T \min(t,s)\, dt\, ds = \int_0^T \left(\int_0^s t\, dt + \int_s^T s \,dt\right) ds $$
user957
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Finding $\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}$

The original question was to evaluate: $$\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx$$ Using the substitution $x=\tan^2 \theta $, I simplified it down to the integral $\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}$. From…
Trogdor
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Some clever trick is required for this Integral with irrational power of cosine as integrand.

See this: $$\newcommand{\b}[1]{\left(#1\right)}\left\lfloor\frac{\displaystyle\int_0^{\pi/2}\cos^{\sqrt{13}-1}x{\rm d}x}{\displaystyle\int_0^{\pi/2}\cos^{\sqrt{13}+1}x{\rm d}x}\right\rfloor$$ Well I could only think of Cauchy-Schwarz, but it is also…
RE60K
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How do I calculate the triple integral $\int_{0}^{2\pi}\int_0^1 \int_0^1 xy\sqrt{x^2 + y^2 -2xy\cos(\theta)} dx \text{ }dy \text{ } d\theta$?

I have tried a couple substitutions which didn't pan out, and I tried differentiating under the integral sign on $$ I(\theta) = \int_0^1 \int_0^1 xy\sqrt{x^2 + y^2 -2xy\cos(\theta)} dx \text{ }dy $$ to get $$ \frac{dI}{d\theta} = -\sin(\theta)…
CameronJWhitehead
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Change of variable in integration causes upper and lower unequal limits to be same and makes finite integral zero.

Let $t=1+\sin x$ in the following integral, then we encounter something weird!! Note: $t(x)=1+\sin x$, so $t(0)=1+\sin0=1+0=1$ and $t(\pi)=1+\sin\pi=1+0=1$ $$2=\int_0^{\pi}\frac{{\rm d}x}{1+\sin x}=\int_{1}^{1}(*){\rm d}t=0??$$ It was actually asked…
RE60K
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Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$

$$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$ How to integrate? I tried the substitution $x=\sin\theta$ but didn't work.
Parvatinandan
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