Question

Evaluate the integral: \[ \int_0^{\frac{\pi}{2}} \frac{\sin \left( \frac{\pi}{4} + x \right) + \sin \left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} \, dx \]

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

Hint:

Use trigonometric identities to simplify the sum of sines in the numerator and the expression in the denominator before evaluating the integral.

Answer 1

The Correct Option is B

We are given the integral: \[ \int_0^{\frac{\pi}{2}} \frac{\sin \left( \frac{\pi}{4} + x \right) + \sin \left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} \, dx \] Step 1: Use the sum of sines identity \( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \) to simplify the numerator. Step 2: Simplify the denominator using the identity \( \cos x + \sin x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \). Step 3: The integral becomes a simpler trigonometric integral, and after solving, we find the result: \[ \frac{\pi}{2\sqrt{2}} \] % Final Answer The value of the integral is \( \frac{\pi}{2\sqrt{2}} \).

Answer 2

Step 1: Write the integral clearly
\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin \left( \frac{\pi}{4} + x \right) + \sin \left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} \, dx \]

Step 2: Use the sum-to-product identity for the numerator
Recall that:
\[ \sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \]
Apply it with \(A = \frac{\pi}{4} + x\) and \(B = \frac{3\pi}{4} + x\):
\[ \sin \left( \frac{\pi}{4} + x \right) + \sin \left( \frac{3\pi}{4} + x \right) = 2 \sin \left( \frac{\frac{\pi}{4} + x + \frac{3\pi}{4} + x}{2} \right) \cos \left( \frac{\frac{\pi}{4} + x - \left(\frac{3\pi}{4} + x\right)}{2} \right) \]
Simplify inside the functions:
\[ = 2 \sin \left( \frac{\pi + 2x}{2} \right) \cos \left( \frac{-\pi/2}{2} \right) = 2 \sin \left( \frac{\pi}{2} + x \right) \cos \left( -\frac{\pi}{4} \right) \]
Since \(\cos (-\theta) = \cos \theta\),
\[ = 2 \sin \left( \frac{\pi}{2} + x \right) \cos \frac{\pi}{4} = 2 \sin \left( \frac{\pi}{2} + x \right) \times \frac{1}{\sqrt{2}} = \sqrt{2} \sin \left( \frac{\pi}{2} + x \right) \]

Step 3: Simplify \(\sin \left( \frac{\pi}{2} + x \right)\)
\[ \sin \left( \frac{\pi}{2} + x \right) = \cos x \]

Step 4: Rewrite the integral
\[ I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2} \cos x}{\cos x + \sin x} \, dx = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} \, dx \]

Step 5: Evaluate the integral inside
Let:
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} \, dx \]
Rewrite denominator:
\[ \cos x + \sin x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \]
But here, instead of substitution, consider:
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x - \sin x}{\cos x + \sin x} \, dx = \int_0^{\frac{\pi}{2}} \left( 1 - \frac{\sin x}{\cos x + \sin x} \right) dx \]
So,
\[ I_1 = \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \frac{\sin x}{\cos x + \sin x} \, dx = \frac{\pi}{2} - I_2 \]
where
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\cos x + \sin x} \, dx \]

Step 6: Note the symmetry
By symmetry,
\[ I_1 = I_2 \]
because if we replace \(x\) by \(\frac{\pi}{2} - x\), \(\cos x\) and \(\sin x\) interchange.

Step 7: Use \(I_1 + I_2 = \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\cos x + \sin x} dx = \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}\)
Since \(I_1 = I_2\),
\[ 2 I_1 = \frac{\pi}{2} \implies I_1 = \frac{\pi}{4} \]

Step 8: Calculate the original integral
\[ I = \sqrt{2} \times I_1 = \sqrt{2} \times \frac{\pi}{4} = \frac{\pi}{2 \sqrt{2}} \]

Final Answer:
\[ \int_0^{\frac{\pi}{2}} \frac{\sin \left( \frac{\pi}{4} + x \right) + \sin \left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} \, dx = \frac{\pi}{2 \sqrt{2}} \]