Question

$\Gamma(n) = \int_0^{\infty} e^{-x} x^{n-1} \, dx, (n > 0)$ , then the value of $\int_0^{\pi/2} \cos^n x \, dx$ is given by}

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

Hint:

Always remember the identity $\Gamma(1/2) = \sqrt{\pi}$. This value appears as a multiplier in many definite integrals involving circular functions evaluated over $[0, \pi/2]$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem uses the relationship between the Beta and Gamma functions to evaluate trigonometric definite integrals.
Key Formula or Approach:
The standard integral identity using the Beta function is: \[ \int_0^{\pi/2} \sin^p x \cos^q x \, dx = \frac{1}{2} B\left( \frac{p+1}{2}, \frac{q+1}{2} \right) \] And the relationship between Beta and Gamma functions is: \[ B(u, v) = \frac{\Gamma(u) \Gamma(v)}{\Gamma(u+v)} \]

Step 2: Detailed Explanation:
Let the given integral be $I$: \[ I = \int_0^{\pi/2} \cos^n x \, dx \]
We can rewrite this integral by introducing a sine term with a power of 0: \[ I = \int_0^{\pi/2} \sin^0 x \cos^n x \, dx \]
Here, $p = 0$ and $q = n$.
Substitute these values into the Beta-Gamma relation: \[ I = \frac{1}{2} B\left( \frac{0+1}{2}, \frac{n+1}{2} \right) = \frac{1}{2} B\left( \frac{1}{2}, \frac{n+1}{2} \right) \] \[ = \frac{1}{2} \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{1}{2} + \frac{n+1}{2}\right)} \] \[ = \frac{1}{2} \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} \]
Since $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$, we substitute this value: \[ I = \frac{\Gamma\left(\frac{n+1}{2}\right) \sqrt{\pi}}{2 \Gamma\left(\frac{n+2}{2}\right)} \]

Step 3: Final Answer:
The value of the definite integral is $\frac{\Gamma\left(\frac{n+1}{2}\right) \sqrt{\pi}}{2 \Gamma\left(\frac{n+2}{2}\right)}$.