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:

$$I = 4\ln3$$