nyuuzyou/wikis huggingface dataset:
Definite Integral from 0 to 1 of x^4 (1 - x)^4 over (1 + x^2)

== Theorem ==
== Proof ==
We then establish a reduction formula:
Then it is a matter of evaluating the individual integrals.
It remains to gather up the terms.
First the terms in $x^5$ and $x^7$, as there is some important cancelling out:
Thus we continue:
The result follows.
== Sources ==
* : $0,001264489 \ldots$
Category:Examples of Definite Integrals
Definite Integral from 0 to 2 Pi of Logarithm of a plus b Cosine x

== Theorem ==
:$\ds \int_0^{2 \pi} \map \ln {a + b \cos x} \rd x = 2 \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
where:
:$b$ is a real number
:$a$ is a positive real number with $a \ge \size b$.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.106.2$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to 2 Pi of Reciprocal of One minus 2 a Cosine x plus a Squared

== Theorem ==
:$\ds \int_0^{2 \pi} \frac {\d x} {1 - 2 a \cos x + a^2} = \frac {2 \pi} {1 - a^2}$
where $a$ is a real number with $0 < a < 1$.
== Proof ==
Let $C$ be the unit open disk centred at $0$. 
The boundary of $C$, $\partial C$, can be parameterized by: 
:$\map \gamma \theta = e^{i \theta}$
for $0 \le \theta \le 2 \pi$. 
We have: 
The integrand has poles: 
:$z_1 = a$
and: 
:$z_2 = \dfrac 1 a$
We have $0 < a < 1$, so $\dfrac 1 a > 1$. 
So:
:$z_2$ lies outside the closed disk $\size z \le 1$
and:
:$z_1$ lies in $C$
Thus the only pole of concern is $z_1$. 
Therefore: 
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.46$
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Cosine x

== Theorem ==
:$\ds \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$
where $a$ and $b$ are real numbers with $a > b > 0$.
== Proof 1 ==
 
== Proof 2 ==
== Also see ==
* Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {\paren {a + b \sin x}^2}$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.46$
Category:Definite Integrals involving Cosine Function
Category:Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Cosine x
Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Cosine x/Proof 1

== Theorem ==
== Proof ==
From Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \cos x}$, we have: 
:$\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$
We have: 
and: 
giving:
:$\ds \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$
Category:Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Cosine x
Definite Integral from 0 to 2 Pi of Reciprocal of Square of a plus b Sine x

== Theorem ==
:$\ds \int_0^{2 \pi} \frac {\d x} {\paren {a + b \sin x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$
where $a$ and $b$ are real numbers with $a > b > 0$.
== Proof ==
From Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \sin x}$, we have: 
:$\ds \int_0^{2 \pi} \frac {\d x} {a + b \sin x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$
We have: 
and: 
giving:
:$\ds \int_0^{2 \pi} \frac {\d x} {\paren {a + b \cos x}^2} = \frac {2 \pi a} {\paren {a^2 - b^2}^{3/2} }$
== Also see ==
* Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {\paren {a + b \cos x}^2}$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.46$
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x

== Theorem ==
:$\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$
where $a$ and $b$ are real numbers with $a > b > 0$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \sin x}$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.44$
Category:Definite Integrals involving Cosine Function
Category:Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x
Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x/Proof 1

== Theorem ==
== Proof ==
Let $C$ be the unit open disk in the complex plane centred at $0$. 
The boundary of $C$, $\partial C$, can be parameterized by: 
:$\map z x = e^{i x}$
for $0 \le x \le 2 \pi$. 
We have:
The integrand has poles where: 
:$\paren {z + \dfrac a b}^2 - \dfrac {a^2} {b^2} + 1 = 0$
That is, where: 
:$\size {z + \dfrac a b} = \dfrac {\sqrt {a^2 - b^2} } b$
So: 
:$z_1 = \dfrac {-a + \sqrt {a^2 - b^2} } b$
and:
:$z_2 = \dfrac {-a - \sqrt {a^2 - b^2} } b$
are the poles of the integrand. 
We have: 
 
So $z_2$ lies outside the closed disk $\size z \le 1$ for all real $a > b > 0$, so is of no concern. 
We now establish a bound on $z_1$. 
As $a > b > 0$, we have that: 
:$a^2 > a^2 - b^2 > 0$
So, from Square Root is Strictly Increasing: 
:$a > \sqrt {a^2 - b^2} > 0$
Multiplying through by $-2 \sqrt {a^2 - b^2}$: 
:$-2 a \sqrt{a^2 - b^2} < 2 \paren {b^2 - a^2} < 0$
This can be rewritten as: 
giving: 
:$\size {-a + \sqrt {a^2 - b^2} } < b$
Therefore: 
:$\size {z_1} = \size {\dfrac {-a + \sqrt {a^2 - b^2} } b} < 1$
So $z_1$ is the only pole of the integrand lying within $C$. 
Therefore: 
Category:Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x
Definite Integral from 0 to Half Pi of Even Power of Cosine x

== Theorem ==
Let $n \in \Z_{> 0}$ be a positive integer.
Then:
:$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.30$
Category:Definite Integral from 0 to Half Pi of Even Power of Cosine x
Category:Reduction Formula for Integral of Power of Cosine
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Half Pi of Even Power of Cosine x/Proof 1

== Theorem == 
== Proof == 
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \int_0^{\frac \pi 2} \cos^{2 k} x \rd x = \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} \dfrac \pi 2$
from which it is to be shown that:
:$\ds \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} } x \rd x = \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} \dfrac \pi 2$
=== Induction Step ===
This is the induction step:
Let $I_k = \ds \int_0^{\frac \pi 2} \cos^{2 k} x \rd x$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{> 0}: \int_0^{\frac \pi 2} \cos^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$
Category:Definite Integral from 0 to Half Pi of Even Power of Cosine x
Definite Integral from 0 to Half Pi of Even Power of Sine x

== Theorem ==
for $n \in \Z_{>0}$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Reduction Formula for Definite Integral of Power of Sine
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.30$
* : Back endpapers: A Brief Table of Integrals: $141$.
* : $\S 17.4 \ (2)$
Category:Definite Integral from 0 to Half Pi of Even Power of Sine x
Category:Reduction Formula for Definite Integral of Power of Sine
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Half Pi of Logarithm of Cosine x

== Theorem ==
:$\ds \int_0^{\pi/2} \map \ln {\cos x} \rd x = -\frac \pi 2 \ln 2$
== Proof ==
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Proof 1 we have: 
:$\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = \int_0^{\pi/2} \map \ln {\cos x} \rd x$
and:
:$\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
The result follows.
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.102$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Half Pi of Logarithm of Sine x

== Theorem ==
:$\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.102$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Sine Function
Category:Definite Integral from 0 to Half Pi of Logarithm of Sine x
Definite Integral from 0 to Half Pi of Logarithm of Sine x/Lemma

== Lemma for Definite Integral from 0 to Half Pi of Logarithm of Sine x ==
:$\ds \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$
== Proof ==
We have: 
We can therefore write: 
Category:Definite Integral from 0 to Half Pi of Logarithm of Sine x
Definite Integral from 0 to Half Pi of Logarithm of Sine x/Proof 1

== Theorem ==
== Proof ==
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma, we have: 
:$\ds \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$
We also have: 
giving: 
Therefore: 
:$\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
Category:Definite Integral from 0 to Half Pi of Logarithm of Sine x
Definite Integral from 0 to Half Pi of Logarithm of Sine x/Proof 2

== Theorem ==
== Proof ==
By , we have: 
:$\ds \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$
By Product of Sines of Fractions of Pi, we have: 
:$\ds \prod_{k \mathop = 1}^{n - 1} \map \sin {\frac {k \pi} n} = \frac n {2^{n - 1} }$
Therefore, we have: 
We have: 
We can show the first term to vanish: 
So: 
giving:
:$\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
== Sources ==
* 
Category:Definite Integral from 0 to Half Pi of Logarithm of Sine x
Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx

== Theorem ==
For $n \in \N_{>0}$:
:$\ds \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \rd x = -\frac \pi {4 n}$
== Proof ==
First we have:
hence $\ds \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x = 0$.
Thus:
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Half Pi of Odd Power of Cosine x

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
== Proof 1 == 
== Proof 2 ==
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.31$
Category:Definite Integral from 0 to Half Pi of Odd Power of Cosine x
Category:Reduction Formula for Integral of Power of Cosine
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Half Pi of Odd Power of Cosine x/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x = \dfrac {\paren {2^k k!}^2} {\paren {2 k + 1}!}$
from which it is to be shown that:
:$\ds \int_0^{\frac \pi 2} \cos^{2 \paren {k + 1} + 1} x \rd x = \dfrac {\paren {2^{k + 1} \paren {k + 1}!}^2} {\paren {2 \paren {k + 1} + 1}!}$
=== Induction Step ===
This is the induction step:
Let $I_k = \ds \int_0^{\frac \pi 2} \cos^{2 k + 1} x \rd x$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: \int_0^{\frac \pi 2} \cos^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
Category:Definite Integral from 0 to Half Pi of Odd Power of Cosine x
Definite Integral from 0 to Half Pi of Odd Power of Sine x

== Theorem ==
for $n \in \Z_{>0}$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Reduction Formula for Definite Integral of Power of Sine
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.31$
* : Back endpapers: A Brief Table of Integrals: $141$.
* : $\S 17.4 \ (2)$
Category:Definite Integral from 0 to Half Pi of Odd Power of Sine x
Category:Reduction Formula for Definite Integral of Power of Sine
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Half Pi of Reciprocal of One plus Power of Tan x

== Theorem ==
:$\ds \int_0^{\pi/2} \frac {\d x} {1 + \tan^m x} = \frac \pi 4$
where $m$ is a real number.
== Proof ==
So:
giving: 
:$\ds \int_0^{\pi/2} \frac {\d x} {1 + \tan^m x} = \frac \pi 4$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.61$
Category:Definite Integrals involving Tangent Function
Definite Integral from 0 to Half Pi of Reciprocal of a plus b Cosine x

== Theorem ==
:$\ds \int_0^{\pi/2} \frac 1 {a + b \cos x} \rd x = \frac 1 {\sqrt {a^2 - b^2} } \map \arccos {\frac b a}$
where $a$ and $b$ are real numbers with $a > b > 0$. 
== Proof ==
Since $a > b > 0$, we have $a^2 > b^2$.
So:
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.45$
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Half Pi of Sine x by Logarithm of Sine x

== Theorem ==
:$\ds \int_0^{\pi/2} \sin x \map \ln {\sin x} \rd x = \ln 2 - 1$
== Proof ==
It remains to compute: 
:$\ds \lim_{x \mathop \to 0^+} \paren {\map \ln {\sin x} - \map \ln {\tan \frac x 2} }$
We have:
giving: 
:$\ds \int_0^{\pi/2} \sin x \map \ln {\sin x} \rd x = \ln 2 - 1$
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.105$
Category:Definite Integrals involving Sine Function
Category:Definite Integrals involving Logarithm Function
Definite Integral from 0 to Half Pi of Square of Cosine x

== Theorem ==
:$\ds \int_0^{\frac \pi 2} \cos^2 x \rd x = \frac \pi 4$
== Proof 1 ==
 
== Proof 2 ==
== Also see ==
* Definite Integral from $0$ to $\dfrac \pi 2$ of $\sin^2 x$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.29$
Category:Definite Integrals involving Cosine Function
Category:Definite Integral from 0 to Half Pi of Square of Cosine x
Definite Integral from 0 to Half Pi of Square of Cosine x/Proof 2

== Theorem ==
== Proof ==
We have: 
So:
giving:
:$\ds \int_0^{\frac \pi 2} \cos^2 x \rd x = \frac \pi 4$
Category:Definite Integral from 0 to Half Pi of Square of Cosine x
Definite Integral from 0 to Half Pi of Square of Logarithm of Cosine x

== Theorem ==
:$\ds \int_0^{\pi/2} \paren {\map \ln {\cos x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.103$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Half Pi of Square of Logarithm of Sine x

== Theorem ==
:$\ds \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$
== Proof ==
From Fourier Series for $\map \ln {\sin x}$ from $0$ to $\pi$:
:$\ds \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$
Then, by Parseval's Theorem:
We then have: 
So: 
:$\ds \frac 4 \pi \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = 2 \paren {\ln 2}^2 + \frac {\pi^2} 6$
Multiplying by $\dfrac \pi 4$: 
:$\ds \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.103$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Half Pi of Square of Sine x

== Theorem ==
:$\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$
== Proof 1 ==
 
== Proof 2 ==
== Also see ==
* Definite Integral from $0$ to $\dfrac \pi 2$ of $\cos^2 x$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.29$
Category:Definite Integrals involving Sine Function
Category:Definite Integral from 0 to Half Pi of Square of Sine x
Definite Integral from 0 to Half Pi of Square of Sine x/Proof 2

== Theorem ==
== Proof ==
We have: 
So:
giving:
:$\ds \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$
Category:Definite Integral from 0 to Half Pi of Square of Sine x
Definite Integral from 0 to Half Pi of x over Sine x

== Theorem ==
:$\ds \int_0^{\pi/2} \frac x {\sin x} \rd x = 2 G$
 
where $G$ is Catalan's constant.
== Proof ==
From Definite Integral from $0$ to $1$ of $\dfrac {\arctan x} x$, we have: 
:$\ds \int_0^1 \frac {\arctan x} x \rd x = G$
Let: 
:$x = \tan \theta$ 
By Derivative of Tangent Function, we have: 
:$\ds \frac {\d x} {\d \theta} = \sec^2 \theta$
We have, by Arctangent of Zero is Zero: 
:as $x \to 0$, $\theta \to 0$.
We also have, by Arctangent of One:
:as $x \to 1$, $\theta \to \dfrac \pi 4$
We therefore have: 
giving: 
:$\ds \int_0^{\pi/2} \frac \phi {\sin \phi} \rd \phi = 2 G$
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.62$
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Infinity of x^m over (x^n + a^n)^r

== Theorem ==
:$\ds \int_0^\infty \frac {x^m \rd x} {\paren {a^n + x^n}^r} = \frac {\paren {-1}^{r - 1} \pi a^{m + 1 - n r} \, \map \Gamma {\frac {m + 1} n} } {n \sin \frac {\paren {m + 1} \pi} n \paren {r - 1}! \, \map \Gamma {\frac {m + 1} n - r + 1} }$
for:
:$0 < m + 1 < n r$
:$r + 1 - \dfrac {m + 1} n \notin \N_{>0}$
:$r \in \N_{>0}$.
== Proof ==
== Sources ==
* : $\S 15$: Definite Integrals involving Rational or Irrational expressions: $15.25$
Category:Examples of Definite Integrals
Definite Integral from 0 to Pi of Cosine of m x by Cosine of n x

== Theorem ==
Let $m, n \in \Z$ be integers.
Then:
:$\ds \int_0^\pi \cos m x \cos n x \rd x = \begin {cases} 0 & : m \ne n \\ \dfrac \pi 2 & : m = n \end {cases}$
That is:
:$\ds \int_0^\pi \cos m x \cos n x \rd x = \dfrac \pi 2 \delta_{m n}$
where $\delta_{m n}$ is the Kronecker delta.
== Proof ==
Let $m \ne n$.
When $m = n$ we have:
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.27$
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Pi of Logarithm of a plus b Cosine x

== Theorem ==
:$\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
where:
:$b$ is a real number
:$a$ is a positive real number with $a \ge \size b$.
== Proof ==
Fix $b \in \R$ and define: 
:$\ds \map I a = \int_0^\pi \map \ln {a + b \cos x} \rd x$
for $a \ge \size b$.
We have: 
So, by Primitive of $\sqrt {x^2 - a^2}$: Logarithm Form: 
:$\map I a = \pi \map \ln {a + \sqrt {a^2 - b^2} } + C$
for all $a \ge \size b$, for some $C \in \R$.
We now investigate cases according to the sign of $b$. 
Let $b = 0$.
We have: 
On the other hand: 
So, by Sum of Logarithms:
:$\pi \ln 2 + \pi \ln a + C = \pi \ln a$
so:
:$C = -\pi \ln 2$
giving: 
in the case $b = 0$.
Let $b > 0$.
Then: 
On the other hand:
giving: 
:$C = -\pi \ln 2$
So we have: 
:$\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
in the case $b > 0$ too.
Let $b < 0$.
We have: 
On the other hand:
We have in this case: 
:$C = -\pi \ln 2$
So we have: 
:$\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
in the case $b < 0$ too.
The result follows.
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.107$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Pi of Sec x by Logarithm of One plus b Cosine x over One plus a Cosine x

== Theorem ==
:$\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$
where $a$ and $b$ are real numbers with $-1 < a, b < 1$.
== Proof ==
Note that by Difference of Logarithms: 
:$\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \int_0^{\pi/2} \sec x \map \ln {1 + b \cos x} \rd x - \int_0^{\pi/2} \sec \map \ln {1 + a \cos x} \rd x$
For each $\alpha \in \openint {-1} 1$, set: 
:$\ds \map I \alpha = \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x$
Then: 
:$\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \map I b - \map I a$
We have: 
Then:
From Derivative of Arccosine Function, we have: 
:$\ds \frac \d {\d \alpha} \arccos \alpha = -\frac 1 {\sqrt {1 - \alpha^2} }$
so:
so:
:$\ds \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$
as required.
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.110$
Category:Definite Integrals involving Cosine Function
Category:Definite Integrals involving Logarithm Function
Definite Integral from 0 to Pi of Sine of m x by Cosine of n x

== Theorem ==
Let $m, n \in \Z$ be integers.
Then:
:$\ds \int_0^\pi \sin m x \cos n x \rd x = \begin{cases} 0 & : m + n \text { even} \\ \dfrac {2 m} {m^2 - n^2} & : m + n \text { odd} \end{cases}$
== Proof ==
First we address the special case where $m = n$.
In this case $m + n = m + m = 2 m$ is even.
We have:
So in this case, $\ds \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even.
Let $m \ne n$.
When $m + n$ is an even integer, we have:
This shows that in all cases $\ds \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even.
When $m + n$ is an odd integer, we have:
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.28$
Category:Definite Integrals involving Cosine Function
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Pi of Sine of m x by Sine of n x

== Theorem ==
Let $m, n \in \Z$ be integers.
Then:
:$\ds \int_0^\pi \sin m x \sin n x \rd x = \begin{cases} 0 & : m \ne n \\ \dfrac \pi 2 & : m = n \end{cases}$
That is:
:$\ds \int_0^\pi \sin m x \sin n x \rd x = \dfrac \pi 2 \delta_{m n}$
where $\delta_{m n}$ is the Kronecker delta.
== Proof ==
Let $m \ne n$.
When $m = n$ we have:
== Sources ==
* : $\S 15$: Definite Integrals involving Trigonometric Functions: $15.26$
Category:Definite Integrals involving Sine Function
Definite Integral from 0 to Pi of a Squared minus 2 a b Cosine x plus b Squared

== Theorem ==
:$\ds \int_0^\pi \map \ln {a^2 - 2 a b \cos x + b^2} \rd x = \begin{cases}2 \pi \ln a & a \ge b > 0 \\ 2 \pi \ln b & b \ge a > 0\end{cases}$
== Proof ==
Note that: 
:$\paren {a - b}^2 \ge 0$
so by Square of Sum:
:$a^2 - 2 a b + b^2 \ge 0$
So:
:$a^2 + b^2 \ge 2 a b = \size {-2 a b}$
so we may apply Definite Integral from $0$ to $\pi$ of $\map \ln {a + b \cos x}$.
We then have:
Note that if $a \ge b > 0$ we have:
Note that if $b \ge a > 0$ we have: 
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.108$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Cosine Function
Definite Integral from 0 to Pi of x by Logarithm of Sine x

== Theorem ==
:$\ds \int_0^\pi x \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$
== Proof ==
So: 
giving: 
:$\ds \int_0^\pi x \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$
== Sources ==
* : $\S 15$: Definite Integrals involving Logarithmic Functions: $15.104$
Category:Definite Integrals involving Logarithm Function
Category:Definite Integrals involving Sine Function