Question

The value of \( \int_0^{\frac{\pi}{2}} \frac{\sin\left( \frac{\pi}{4} + x \right) + \sin\left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} \, dx \) is:

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

Hint:

When dealing with integrals involving trigonometric sums, use sum-to-product identities to simplify the numerator.

Answer 1

The Correct Option is B

Step 1: Let \( I = \int_0^{\frac{\pi}{2}} \frac{\sin\left( \frac{\pi}{4} + x \right) + \sin\left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} dx \). We will simplify the integral using trigonometric identities. \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin\left( \frac{\pi}{4} + x \right) + \sin\left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} dx \] Apply sum-to-product identities to the numerator: \[ I = \int_0^{\frac{\pi}{2}} \frac{2\sin\left(\frac{\pi}{2} + x \right)}{\cos x + \sin x} dx \] Step 2: Simplify further using the identity \( \sin\left(\frac{\pi}{2} + x \right) = \cos x \), and we get: \[ I = \int_0^{\frac{\pi}{2}} \frac{2\cos x}{\cos x + \sin x} dx \] Now, we can use substitution and trigonometric simplifications to arrive at the solution. The value of the integral is \( \frac{\pi}{2\sqrt{2}} \).

Answer 2

Step 1: Simplify the given expression
We are asked to evaluate the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin\left( \frac{\pi}{4} + x \right) + \sin\left( \frac{3\pi}{4} + x \right)}{\cos x + \sin x} \, dx \] Using the trigonometric identity for the sum of sines: \[ \sin(A + x) = \sin A \cos x + \cos A \sin x \] We can rewrite the two sine terms: \[ \sin\left( \frac{\pi}{4} + x \right) = \sin\frac{\pi}{4} \cos x + \cos\frac{\pi}{4} \sin x = \frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x \] \[ \sin\left( \frac{3\pi}{4} + x \right) = \sin\frac{3\pi}{4} \cos x + \cos\frac{3\pi}{4} \sin x = \frac{\sqrt{2}}{2} \cos x - \frac{\sqrt{2}}{2} \sin x \] Now, add these two sine terms together: \[ \sin\left( \frac{\pi}{4} + x \right) + \sin\left( \frac{3\pi}{4} + x \right) = \left( \frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x \right) + \left( \frac{\sqrt{2}}{2} \cos x - \frac{\sqrt{2}}{2} \sin x \right) \] \[ = \sqrt{2} \cos x \] Thus, the integral simplifies to: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2} \cos x}{\cos x + \sin x} \, dx \]
Step 2: Simplify the denominator
Next, we simplify the denominator \( \cos x + \sin x \). Notice that: \[ \cos x + \sin x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) \] This is equivalent to: \[ \cos x + \sin x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right) \] So the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2} \cos x}{\sqrt{2} \cos \left( x - \frac{\pi}{4} \right)} \, dx = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos \left( x - \frac{\pi}{4} \right)} \, dx \]
Step 3: Use a substitution to simplify further
Let’s perform the substitution: \[ u = x - \frac{\pi}{4}, \quad du = dx \] When \( x = 0 \), \( u = -\frac{\pi}{4} \), and when \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{4} \). The integral becomes: \[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos\left( u + \frac{\pi}{4} \right)}{\cos u} \, du \] Now, use the identity: \[ \cos\left( u + \frac{\pi}{4} \right) = \cos u \cos \frac{\pi}{4} - \sin u \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \cos u - \frac{\sqrt{2}}{2} \sin u \] Thus, the integral becomes: \[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\sqrt{2}}{2} \cos u - \frac{\sqrt{2}}{2} \sin u}{\cos u} \, du \] \[ I = \frac{\sqrt{2}}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( 1 - \tan u \right) du \]
Step 4: Solve the integral
Now, split the integral: \[ I = \frac{\sqrt{2}}{2} \left[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, du - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan u \, du \right] \] The first integral: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, du = \frac{\pi}{2} \] The second integral: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan u \, du = 0 \] (since \(\tan u\) is an odd function and the limits are symmetric). Thus, the value of the integral is: \[ I = \frac{\sqrt{2}}{2} \times \frac{\pi}{2} = \frac{\pi}{2\sqrt{2}} \]
Step 5: Conclusion
The value of the integral is: \[ \boxed{\frac{\pi}{2\sqrt{2}}} \]