Question

If \(\Gamma(n)=\int_{0}^{\infty} e^{-x}x^{n-1}\,dx,\; n > 0\), then the value of \(\int_{0}^{\pi/2}\sin^{n}x\,dx\) is given by:

• A. $\frac{\Gamma\left(\frac{\text{n}+1}{2}\right)}{\Gamma\left(\frac{\text{n}+2}{2}\right)} \frac{\sqrt{\pi}}{2}$
• B. $\frac{\Gamma\left(\frac{\text{n}}{2}\right)}{\Gamma\left(\frac{\text{n}+1}{2}\right)} \frac{\sqrt{\pi}}{2}$
• C. $\frac{\Gamma\left(\frac{\text{n}+2}{2}\right)}{\Gamma\left(\frac{\text{n}+1}{2}\right)} \frac{\sqrt{\pi}}{2}$
• D. $\frac{\Gamma(\text{n})}{\Gamma(\text{n}+1)} \frac{\sqrt{\pi}}{2}$

Hint:

This trigonometric integral is related to Wallis' formula. Using the Beta-Gamma function shortcut is the most robust way to solve such definite integrals in competitive exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Definite integrals of the form $\int_{0}^{\pi/2} \sin^p x \cos^q x \, dx$ can be evaluated elegantly using the Beta and Gamma functions.
Key Formula or Approach:
The relationship between trigonometric integrals and the Gamma function is: \[ \int_{0}^{\pi/2} \sin^p x \cos^q x \, dx = \frac{\Gamma\left(\frac{p+1}{2}\right) \Gamma\left(\frac{q+1}{2}\right)}{2 \Gamma\left(\frac{p+q+2}{2}\right)} \]

Step 2: Detailed Explanation:
The given integral is: \[ I = \int_{0}^{\pi/2} \sin^n x \, dx \] We can write this as: \[ I = \int_{0}^{\pi/2} \sin^n x \cos^0 x \, dx \] Applying the formula with $p = n$ and $q = 0$: \[ I = \frac{\Gamma\left(\frac{n+1}{2}\right) \Gamma\left(\frac{0+1}{2}\right)}{2 \Gamma\left(\frac{n+0+2}{2}\right)} \] \[ I = \frac{\Gamma\left(\frac{n+1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2 \Gamma\left(\frac{n+2}{2}\right)} \] We know that the value of $\Gamma\left(\frac{1}{2}\right)$ is: \[ \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \] Substitute this value back into the expression: \[ I = \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} \frac{\sqrt{\pi}}{2} \] This is the required value of the definite integral.

Step 3: Final Answer:
The value matches Option (A).