Question

The integral \[ 80 \int_0^{\frac{\pi}{4}} \frac{(\sin \theta + \cos \theta)}{(9 + 16 \sin 2\theta)} \, d\theta \] is equal to:

• A. \( 6 \log 4 \)
• B. \( 2 \log 3 \)
• C. \( 4 \log 3 \)
• D. \( 3 \log 4 \)

Hint:

To solve integrals involving trigonometric functions, consider using substitution and applying relevant trigonometric identities to simplify the expression.

Answer 1

The Correct Option is D

Problem:

Evaluate the integral:

\[ I = 80 \int_0^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16\sin 2\theta} \, d\theta \]

Solution:

Apply the identity:

\[ \sin 2\theta = 2\sin\theta \cos\theta = 1 - (\sin\theta - \cos\theta)^2 \]

Substitute:

\[ t = \sin\theta - \cos\theta \quad \Rightarrow \quad dt = (\cos\theta + \sin\theta) d\theta \]

Update the integration limits: \( \theta = 0 \Rightarrow t = -1 \) and \( \theta = \frac{\pi}{4} \Rightarrow t = 0 \)

The integral transforms to:

\[ I = 80 \int_{-1}^{0} \frac{1}{25 - 16t^2} \, dt = \frac{80}{16} \int_{-1}^{0} \frac{1}{\left( \frac{5}{4} \right)^2 - t^2} dt \]

Utilize the standard integral formula:

\[ \int \frac{1}{a^2 - t^2} dt = \frac{1}{2a} \ln\left| \frac{a + t}{a - t} \right| \quad \text{with } a = \frac{5}{4} \]

Compute the definite integral:

\[ I = 5 \cdot \left[ \frac{1}{2 \cdot \frac{5}{4}} \ln\left( \frac{a + t}{a - t} \right) \right]_{-1}^{0} = 5 \cdot \frac{4}{10} \left[ \ln(1) + \ln(3) \right] = 4 \ln 3 \]

Final Answer: \( \boxed{4 \log 3} \)