Mathematics

Questions tagged [trigonometric-integrals]

Relating to integrations consisting of only(mainly) trigonometric functions and/or requiring substitutions by/of trigonometric functions.

1859 questions
60
votes
9 answers

How to evaluate $\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$

Find the value of $$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$$ We have the information that $$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$
math110
  • 95,328
  • 17
  • 156
  • 531
45
votes
11 answers

Integral $\int_0^\infty \frac{\tanh^2x}{x^2}dx$

It appears that $$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$ (so far I have about $1000$ decimal digits to confirm that). After changing variable $x=-\tfrac12\ln z$ , it takes an equivalent form…
Vladimir Reshetnikov
  • 32,880
41
votes
16 answers

Is there another way to solve $\int_{0}^{\pi}\frac{1-\sin x}{\sin x+1}dx$

My way to solve this integral. I wonder is there another way to solve it as it's very long for me. $$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$ Let $$u=\tan (\frac{x}{2})$$ $$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$ By Weierstrass Substitution…
Mathxx
  • 8,470
38
votes
5 answers

Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$

I came across the improper trigonometric integral recently shared in an on-line forum \begin{align} \int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\ \end{align} which amuses me because of its appearance. What is more amusing…
Quanto
  • 125,449
38
votes
4 answers

Bizarre Integral $\int_0^1 \frac{\tan^{-1}{\frac{88\sqrt{21}}{215+36x^2}}}{\sqrt{1-x^2}} {d}x = \frac{\pi^2}{6}$

Does the following equality hold? $$\int_0^1 \frac{\tan^{-1}{\frac{88\sqrt{21}}{215+36x^2}}}{\sqrt{1-x^2}} \, \text{d}x = \frac{\pi^2}{6}$$ The supposed equality holds to 61 decimal places in Mathematica, which fails to numerically evaluate it after…
Jack Tiger Lam
  • 4,724
31
votes
10 answers

Why $ \int_0^{2\pi}\frac{\cos t - r}{1 - 2r\cos t + r^2}\,dt=0$

Define $$ I(r) = \int_0^{2\pi}\frac{\cos t- r}{1 - 2r\cos t + r^2}\,dt $$ Numerical experiments suggest that $I(r) = 0$ for all $r\in [0,1)$ . But I can't show this analytically. This integral appears when computing the Cauchy transform of…
amsmath
  • 9,999
30
votes
1 answer

Prove $\int_0^{\sqrt2/4}\frac1{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)(x-1+x\sqrt{9-16x})}{1-2x}}dx=\frac{\pi^2}{8}$ (from a probability question)

Let $$I=\int_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)(x-1+x\sqrt{9-16x})}{1-2x}}dx$$ Prove that $I=\dfrac{\pi^2}{8}$ . Wolfram suggests that it's true but does not find the antiderivative . Here is the graph of the function being…
Dan
  • 41,295
30
votes
10 answers

Various ways to calculate $\int \sin(x) \cos(x) \, \mathrm{d}x$

Consider the integral $$\mathcal{I} := \int \sin(x) \cos(x) \, \mathrm{d} x$$ $ \newcommand{\II}{\mathcal{I}} \newcommand{\d}{\mathrm{d}} $ This is one of my favorite basic integrals to think about as an instructor, because on the face of it, there…
PrincessEev
  • 51,605
30
votes
1 answer

Evaluating $\left.\int_0^{\pi/2}\sqrt{1+\frac1{\sqrt{1+\tan^nx}}}\text dx \middle/\int_0^{\pi/2}\sqrt{1-\frac1{\sqrt{1+\tan^nx}}}\text dx\right.$

Evaluate the integral ratio $$\dfrac{I_1}{I_2}=\dfrac{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}$$ Using…
V.G
  • 4,302
29
votes
4 answers

Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$

Please help me to evaluate this integral: $$\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$$ Using substitution $x=2\arctan t$ it can be transformed to: $$\int_0^\infty\frac{2}{1+t^2}\arctan^2\left(\frac{2t}{3+t^2}\right)dt$$ Then I tried…
Marty Colos
  • 1
29
votes
7 answers

How to evaluate $I=\int\limits_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$

Prima facie, this integral seems easy to calculate,but alas, this not's case $$I=\int\limits_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$$ The numerical value is I=-1.122690024730644497584272... How to evaluate this integral? By against,I find:…
user178256
  • 5,861
28
votes
6 answers

An interesting trigonometric integral

Yesterday I took part in an entry competition to one of the MSc programs in my university, and during one of the mathematical tests, I had to solve some integral and differential equations. One of those integrals had the following trigonometric…
o.spectrum
  • 1,411
  • 1
  • 12
  • 22
27
votes
2 answers

Show that $\int_0^{\pi/3}\arccos^2(2\sin^2 x-\cos x)\mathrm dx=\frac{19\pi^3}{135}$

Show that $$\int_0^{\pi/3}\arccos^2(2\sin^2 x-\cos x)\mathrm dx=\frac{19\pi^3}{135}$$ Wolfram strongly suggests that it's true. Here is the graph of $y=\arccos^2(2\sin^2 x-\cos x)$ . Context I stumbled upon this when I was trying to answer a similar…
Dan
  • 41,295
26
votes
2 answers

More direct proof that $ \int_0^\infty\frac{\tanh(a x)}{\tanh(\pi x)}\sin(2ax^2)\,dx=\frac{1}{4}$ is independent of parameter $a$

I'm looking for alternative ways to calculate the integral $$ \int_0^\infty\frac{\tanh(\alpha x)}{\tanh(\pi x)}\sin(2\alpha x^2)\,dx=\frac{1}{4},\qquad \alpha>0.\tag {*} $$ It was derived in a lengthy calculation using roundabout method from Fourier…
D.B. Cooper
  • 3,808
26
votes
4 answers

Twist on log of sine and cosine integral $\int_{0}^{\frac{\pi}{2}}x\ln(\sin x)\ln(\cos x)dx$

I ran across this integral and have not been able to evaluate it. $$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$$ I had some ideas. Perhaps some how arrive at…
Cody
  • 14,538
1
2 3
…
99 100