The relation
Before you start: read the Gamma function and Beta function lessons.
\[B(m, n) = \frac{\Gamma(m)\,\Gamma(n)}{\Gamma(m + n)}\]
This one formula turns any Beta function into Gamma values, which you can then work out with \(\Gamma(n + 1) = n\Gamma(n)\) and \(\Gamma\left(\tfrac{1}{2}\right) = \sqrt\pi\). For whole numbers it gives back \(B(m, n) = \dfrac{(m - 1)!\,(n - 1)!}{(m + n - 1)!}\).
Where it comes from
Write \(\Gamma(m)\Gamma(n)\) as a double integral, substitute \(x = r^2\cos^2\theta\), \(y = r^2\sin^2\theta\) (polar coordinates), and the integral separates into a Gamma part in \(r\) and a Beta part in \(\theta\). The full proof uses the double integrals of Unit 5.
Results that follow
Trigonometric integrals (for \(p, q > -1\))
\[\int_0^{\pi/2}\sin^p\theta\,\cos^q\theta\,d\theta = \frac{\Gamma\left(\frac{p + 1}{2}\right)\Gamma\left(\frac{q + 1}{2}\right)}{2\,\Gamma\left(\frac{p + q + 2}{2}\right)}\]
Reflection formula (for \(0 < n < 1\))
\[\Gamma(n)\,\Gamma(1 - n) = \frac{\pi}{\sin n\pi}\]
A useful substitution
\[\int_0^1 x^m(1 - x^n)^p\,dx = \frac{1}{n}\,B\!\left(\frac{m + 1}{n},\; p + 1\right)\]
A quick check of the relation: \(B\left(\tfrac{1}{2}, \tfrac{1}{2}\right) = \dfrac{\Gamma\left(\frac{1}{2}\right)^2}{\Gamma(1)} = \pi\), and directly \(\displaystyle\int_0^1\frac{dx}{\sqrt{x(1 - x)}} = \pi\) too.
Solved examples
Example 1: A trigonometric integral
MediumEvaluate \(\displaystyle\int_0^{\pi/2}\sin^4\theta\cos^2\theta\,d\theta\).
- Here \(p = 4\), \(q = 2\): \(\dfrac{\Gamma\left(\frac{5}{2}\right)\Gamma\left(\frac{3}{2}\right)}{2\,\Gamma(4)}\).
- \(\Gamma\left(\tfrac{5}{2}\right) = \tfrac{3}{4}\sqrt\pi\), \(\Gamma\left(\tfrac{3}{2}\right) = \tfrac{1}{2}\sqrt\pi\), \(\Gamma(4) = 6\).
- \(\dfrac{\frac{3}{8}\pi}{12}\).
Answer\(\dfrac{\pi}{32}\)
Example 2: Square root of tan
Exam levelShow that \(\displaystyle\int_0^{\pi/2}\sqrt{\tan\theta}\,d\theta = \frac{\pi}{\sqrt 2}\).
- \(\sqrt{\tan\theta} = \sin^{1/2}\theta\cos^{-1/2}\theta\), so \(p = \tfrac{1}{2}\), \(q = -\tfrac{1}{2}\).
- The integral is \(\dfrac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{4}\right)}{2\,\Gamma(1)}\).
- Reflection formula with \(n = \tfrac{1}{4}\): \(\Gamma\left(\tfrac{1}{4}\right)\Gamma\left(\tfrac{3}{4}\right) = \dfrac{\pi}{\sin(\pi/4)} = \pi\sqrt 2\).
Answer\(\dfrac{\pi\sqrt 2}{2} = \dfrac{\pi}{\sqrt 2}\)
Example 3: An infinite-range integral
Exam levelEvaluate \(\displaystyle\int_0^\infty\frac{dx}{1 + x^4}\).
- Substitute \(t = x^4\): \(x = t^{1/4}\), \(dx = \tfrac{1}{4}t^{-3/4}\,dt\). The integral becomes \(\dfrac{1}{4}\displaystyle\int_0^\infty\frac{t^{-3/4}}{1 + t}\,dt\).
- In the infinite form of Beta, \(m - 1 = -\tfrac{3}{4}\) gives \(m = \tfrac{1}{4}\), and \(m + n = 1\) gives \(n = \tfrac{3}{4}\). So it is \(\tfrac{1}{4}B\left(\tfrac{1}{4}, \tfrac{3}{4}\right) = \tfrac{1}{4}\Gamma\left(\tfrac{1}{4}\right)\Gamma\left(\tfrac{3}{4}\right)\).
- By the reflection formula, that is \(\tfrac{1}{4}\cdot\pi\sqrt 2\).
Answer\(\dfrac{\pi}{2\sqrt 2}\)
Example 4: Using the substitution formula
MediumEvaluate \(\displaystyle\int_0^1 x^3(1 - x^2)^{5/2}\,dx\).
- Use \(\displaystyle\int_0^1 x^m(1 - x^n)^p\,dx = \frac{1}{n}B\!\left(\frac{m + 1}{n}, p + 1\right)\) with \(m = 3\), \(n = 2\), \(p = \tfrac{5}{2}\): \(\tfrac{1}{2}B\left(2, \tfrac{7}{2}\right)\).
- \(B\left(2, \tfrac{7}{2}\right) = \dfrac{\Gamma(2)\Gamma\left(\frac{7}{2}\right)}{\Gamma\left(\frac{11}{2}\right)} = \dfrac{1}{\frac{9}{2}\cdot\frac{7}{2}} = \dfrac{4}{63}\), using \(\Gamma\left(\tfrac{11}{2}\right) = \tfrac{9}{2}\cdot\tfrac{7}{2}\,\Gamma\left(\tfrac{7}{2}\right)\).
Answer\(\dfrac{2}{63}\)
Example 5: A neat cancellation
Exam levelShow that \(\displaystyle\int_0^\infty\frac{x^8(1 - x^6)}{(1 + x)^{24}}\,dx = 0\).
- Split: \(\displaystyle\int_0^\infty\frac{x^8}{(1 + x)^{24}}\,dx - \int_0^\infty\frac{x^{14}}{(1 + x)^{24}}\,dx\).
- Infinite Beta form: the first is \(B(9, 15)\), the second is \(B(15, 9)\).
- By symmetry \(B(9, 15) = B(15, 9)\).
AnswerThe two terms cancel, so the integral is \(0\).
For any \(\int_0^{\pi/2}\sin^p\theta\cos^q\theta\,d\theta\), the formula works for fractional and even negative powers too, as long as both are greater than \(-1\). That's how \(\sqrt{\tan\theta}\) in Example 2 becomes a two-line answer.
Common mistakes
1. Forgetting the 2 in the trigonometric formula
The denominator is \(2\,\Gamma\left(\frac{p + q + 2}{2}\right)\).
2. Using \(p\) and \(q\) instead of \(\tfrac{p + 1}{2}\) and \(\tfrac{q + 1}{2}\)
The Gamma arguments are \(\tfrac{p + 1}{2}\) and \(\tfrac{q + 1}{2}\), not the powers themselves.
3. Using the reflection formula outside \(0 < n < 1\)
\(\Gamma(n)\Gamma(1 - n) = \dfrac{\pi}{\sin n\pi}\) needs \(n\) between 0 and 1. Otherwise use the recurrence first.
How it's asked in exams
- Evaluate trigonometric integrals such as \(\int_0^{\pi/2}\sin^p\theta\cos^q\theta\,d\theta\), including \(\sqrt{\tan\theta}\) and \(\sqrt{\cot\theta}\).
- Evaluate algebraic or infinite-range integrals by reducing them to Beta form.
- Prove results such as \(\Gamma\left(\tfrac{1}{2}\right) = \sqrt\pi\) or identities between Beta values.
Practice questions
Try each one on paper before you open the answer.
Q1\(\Gamma\left(\tfrac{1}{4}\right)\Gamma\left(\tfrac{3}{4}\right)\)
\(\dfrac{\pi}{\sin(\pi/4)} = \pi\sqrt 2\).
Q2\(\displaystyle\int_0^{\pi/2}\sin^6\theta\,d\theta\)
\(p = 6\), \(q = 0\): \(\dfrac{\Gamma\left(\frac{7}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma(4)} = \dfrac{\frac{15}{8}\pi}{12} = \dfrac{5\pi}{32}\).
Q3\(\displaystyle\int_0^{\pi/2}\sqrt{\cot\theta}\,d\theta\)
Same as Example 2 with \(p\) and \(q\) swapped: \(\dfrac{\pi}{\sqrt 2}\).
Q4\(\displaystyle\int_0^1\frac{dx}{\sqrt{1 - x^4}}\)
\(m = 0\), \(n = 4\), \(p = -\tfrac{1}{2}\): \(\dfrac{1}{4}B\left(\tfrac{1}{4}, \tfrac{1}{2}\right) = \dfrac{\Gamma\left(\frac{1}{4}\right)\sqrt\pi}{4\,\Gamma\left(\frac{3}{4}\right)} \approx 1.311\).
Q5Show that \(B(m + 1, n) = \dfrac{m}{m + n}B(m, n)\).
\(B(m + 1, n) = \dfrac{\Gamma(m + 1)\Gamma(n)}{\Gamma(m + n + 1)} = \dfrac{m\,\Gamma(m)\Gamma(n)}{(m + n)\,\Gamma(m + n)} = \dfrac{m}{m + n}B(m, n)\).
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Frequently asked questions
What is the relation between the Beta and Gamma functions?
\(B(m, n) = \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)}\).
Why is this relation so useful?
Beta integrals are hard to evaluate directly, but Gamma values are easy to compute with \(\Gamma(n + 1) = n\Gamma(n)\) and \(\Gamma\left(\tfrac{1}{2}\right) = \sqrt\pi\).
What is the reflection formula?
\(\Gamma(n)\Gamma(1 - n) = \dfrac{\pi}{\sin n\pi}\) for \(0 < n < 1\). It gives products like \(\Gamma\left(\tfrac{1}{4}\right)\Gamma\left(\tfrac{3}{4}\right) = \pi\sqrt 2\).
Unit 4 complete: what comes next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.