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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How to calculate this definite integral $\int_0^1\frac{{\ln^4 x }}{1+x^2}dx=\frac{5\pi^5}{64}$

$$\displaystyle\int_0^1\dfrac{{\ln^4 x}}{1+x^2}\text{d}x=\dfrac{5\pi^5}{64}$$ let $x=e^{-t}$， $$ \displaystyle\int_0^1\dfrac{({\ln…
SHZ
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if the indefinite integral of $x^x$ was $f(x)$ what would the indefinite integral of $x^{1/x}$ be in terms of $x$ and $f(x)$?

so what this means is if $f(x)$ is the indefinite integral of $x^x dx$ then what would the indefinite integral of $x^{1/x}$ be in terms of $x$ and $f(x)$
Ekadh Singh - Reinstate Monica
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Integrate without expansion?

I want to evaluate $$ \int_0^1 ( 1 - x^2)^{10} dx $$ One way I can do this is by expanding out $(1 - x^2)^{10}$ term by term, but is there a better way to do this?
user680412
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Integral $ \int\limits_{-\infty}^\infty \exp \left[-\frac{(x-x_o)^2}{2 \sigma_x^2}-i (p - p_0) \frac{x}{\hbar}\right] \, dx $

Can somebody show me how to calculate this integral? $$ \int\limits_{-\infty}^\infty \exp \left[-\frac{(x-x_o)^2}{2 \sigma_x^2}-i (p - p_0) \frac{x}{\hbar}\right] \, dx $$ $x_0$, $p_0$, $\hbar$ are constants and $\sigma_x$ is a standard deviation…
71GA
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Finding $\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$

Find $$\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$$ What I tried: $$\sin^4(x)+\cos^4(x)=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x$$ and $$\sin^3 x+\cos^3 x=(\sin x+\cos x)(1-\sin x\cos x)$$ so $$\int\frac{1-\sin^2…
jacky
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Why am I running into problems when I don't fully reduce this integral problem?

I am trying to find the integral of: $$\int \frac{dx}{x\sqrt{x-1}}$$ if I do this $u$-sub: $u = \sqrt{x-1}$ and $u^2 = x-1$ and $x = u^2 + 1$ and $\frac{dx}{du} = 2u$ and $dx = 2udu$ Then I get: $$\int \frac{2udu}{(u^2+1)u}$$ if I do a partial…
Jwan622
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What did I do wrong while trying to solve this integral?

So the question is $$\int x^3\ln(x+1)\,dx$$ and I did it this way: $$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int {x^4\over x+1}\,dx$$ $$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int {x^4-1+1\over x+1}\,dx$$ $$= {1\over 4}\ln(x+1)x^4 - {1\over 4}\int {(x^2…
Ibrahim B.
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Find $\int e^{-x}\cos x\,dx$ without using complex numbers

$\int e^{-x} \cos{x} dx $ - i know how to solve with Euler complex representation, but can't figure out how to solve with integration by parts or something.
Yola
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$\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$???

I saw in a paper by @Jack D'aurizio the following integral $$I=\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$$ Below is my attempt. $$I=4\int_0^{\pi/2}\log^2(\cos x)\mathrm{d}x$$ Then we…
clathratus
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Notation for Higher Antiderivatives?

Higher derivative are blessed with many notations. For example $$ f',f'',...$$ or $$ \frac {dy}{dx}, \frac {d^2 y}{dx^2},...$$ I have not seen any notations for higher anti-derivatives. For example, the higher anti-derivatives of $$f(x)=2x+5$$…
Mohammad Riazi-Kermani
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Please how to find indefinite integral $\int x^{x^x}\mathrm dx$?

Please how to find indefinite integral $$\int x^{x^x}\mathrm dx$$ Thank for any one help me to find it
Ahmed
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Help with $\large \intop\frac{\sqrt{x^2-4}}{x} dx$

I have to $\large \intop\frac{\sqrt{x^2-4}}{x} dx$. Can you tell me what to substitute? Should I substitute $x$ or $\sqrt{x^2-4}$? Would it be better If I'd substitute $x=\tan u$ or should I substitute $\sqrt{x^2-4}= \tan u$?
user60286
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How to know when to complete the square

Question is: $$\int \frac{dx}{ x^2+8x+20}$$ Why can I not just solve for $A/(x+2) +B/(x+10)$ and integrate it this way? The answer on symbolab shows I need to complete the square of the denominator first but I don't know hen to do that or when…
Jessie
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Evaluating a nested log integral

Question:$$\int\limits_0^1\mathrm dx\,\frac {\log\log\frac 1x}{(1+x)^2}=\frac 12\log\frac {\pi}2-\frac {\gamma}2$$ I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation…
Frank
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How to show that this integral is correct?

How can one show that $$\int_0^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha} \tan(x) \tan\left(\frac{x}{2}\right)\right] \sqrt{\sin(x) \tan \left(\frac{x}{2}\right)} \, \mathrm dx=-\frac{\ln(\alpha)}{\sqrt{2}}$$ assume $\alpha\ge1$. I…
user516887
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