Question

The integral is given by:

\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]

is equals to?

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

Hint:

When solving integrals involving trigonometric functions, use standard identities and substitution techniques.

Answer 1

The Correct Option is D

Given integral:

\[ I = 80 \int_0^{\frac{\pi}{2}} \left(\frac{\sin\theta + \cos\theta}{9 + 16(2\sin\theta \cos\theta -1)} \right) d\theta \]

Rewriting the denominator:

\[ I = 80 \int_0^{\frac{\pi}{2}} \frac{\sin\theta + \cos\theta}{9 - 16(1 - 2\sin\theta \cos\theta - 1)} d\theta \]

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

Using substitution:

\(\sin\theta - \cos\theta = t\)

and

\((\cos\theta + \sin\theta) d\theta = dt\)

Thus, the integral simplifies to:

\[ I = 80 \int_{0}^{1} \frac{dt}{25 - 16t^2} \]

Rewriting the denominator:

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

Using the standard integral formula:

\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| \]

Applying limits:

\[ I = \frac{5}{2\left(\frac{5}{4}\right)} \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \Bigg|_{-1}^{0} \]

Evaluating at limits:

\[ I = 2\ln(1) + 4\ln3 \]

Since \( \ln(1) = 0 \), we get:

\[ I = 4\ln3 \]