Advertisements
Advertisements
प्रश्न
Evaluate the following:
`Γ (9/2)`
बेरीज
Advertisements
उत्तर
`Γ (9/2) = (9 /2 - 1) Γ(9/2 - 1)`
= `7/2 Γ 7/2`
= `7/2 5/2 Γ 5/2`
= `7/2 5/2 3/2 Γ 3/2`
= `7/2 5/2 3/2 1/2 Γ (1/2)`
= `(105sqrt(pi))/16`
shaalaa.com
Definite Integrals
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Integral Calculus – 1 - Exercise 2.10 [पृष्ठ ५१]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\]
\[\int\limits_0^1 \log\left( \frac{1}{x} - 1 \right) dx\]
\[\int\limits_3^5 \left( 2 - x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
\[\int\limits_0^{\pi/2} \cos^2 x\ dx .\]
\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]
\[\int\limits_0^1 2^{x - \left[ x \right]} dx\]
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
