Question

$\int_{0}^{\frac{\pi}{2}}\frac{300 \sin x+100 \cos x}{\sin x+\cos x}dx=...$

• A. $100\pi$
• B. $300\pi$
• C. $200\pi$
• D. $150\pi$

Hint:

For $\int_0^{\pi/2} \frac{a \sin x + b \cos x}{\sin x + \cos x} dx$, the result is $\frac{(a+b)}{2} \cdot \frac{\pi}{2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The objective is to evaluate the definite integral of a rational trigonometric function over the interval $[0, \pi/2]$.

Step 2: Key Formula or Approach:
We use the property of definite integrals: \[ \int_0^a f(x) dx = \int_0^a f(a - x) dx \]

Step 3: Detailed Explanation:
Let $I = \int_0^{\pi/2} \frac{300 \sin x + 100 \cos x}{\sin x + \cos x} dx$ -----(1)
Applying the property $x \rightarrow \pi/2 - x$: \[ I = \int_0^{\pi/2} \frac{300 \sin(\frac{\pi}{2} - x) + 100 \cos(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} dx \] \[ I = \int_0^{\pi/2} \frac{300 \cos x + 100 \sin x}{\cos x + \sin x} dx \] -----(2)
Adding equations (1) and (2): \[ 2I = \int_0^{\pi/2} \frac{(300 \sin x + 100 \cos x) + (300 \cos x + 100 \sin x)}{\sin x + \cos x} dx \] \[ 2I = \int_0^{\pi/2} \frac{400(\sin x + \cos x)}{\sin x + \cos x} dx \] \[ 2I = \int_0^{\pi/2} 400 dx \] \[ 2I = [400x]_0^{\pi/2} = 400 \times \frac{\pi}{2} = 200\pi \] \[ I = 100\pi \]

Step 4: Final Answer:
The value of the integral is $100\pi$.