Question

$ \int_{0}^{\frac{\pi}{2}} \frac{\sum_{n=0}^{4} \sin \left( \frac{n\pi}{4} + x \right)}{\cos x + \sin x} dx = $

• A. $ \frac{\pi}{2} $
• B. $ \frac{\pi}{2\sqrt{2}} $
• C. $ \frac{3\pi}{\sqrt{2}} $
• D. $ \frac{(\sqrt{2} + 1)\pi}{4} $

Hint:

Simplify the summation before integrating. Use the property $ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx $ for integrals of the form $ \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx $.

Answer 1

The Correct Option is D

Step 1: Evaluate the summation.
$ \sum_{n=0}^{4} \sin \left( \frac{n\pi}{4} + x \right) = (1 + \sqrt{2}) \cos x $
Step 2: Substitute into the integral.
$ I = (1 + \sqrt{2}) \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx $
Step 3: Evaluate $ J = \int_{0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx $.}
Using $ J = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx $, we get $ 2J = \int_{0}^{\frac{\pi}{2}} 1 dx = \frac{\pi}{2} \implies J = \frac{\pi}{4} $.
Step 4: Calculate $ I $.
$ I = (1 + \sqrt{2}) \frac{\pi}{4} = \frac{(\sqrt{2} + 1)\pi}{4} $
Step 5: Conclusion.
The value of the integral is $ \frac{(\sqrt{2} + 1)\pi}{4} $.