JEE Mathematics 2025 Question with Solution

Given: The integral is

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

Find: Its value and the correct option.

The solution states that the correct option is D.

Using the substitution shown in the solution,

$$\sin\theta - \cos\theta = t$$

so that

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

The working then simplifies the integral to

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

Using the standard result

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

we obtain after applying the limits,

$$I = 2\ln(1) + 4\ln3$$

Since

$$\ln(1) = 0$$

therefore

$$I = 4\ln3$$

This value corresponds to option C from the listed options. However, the solution explicitly marks D as the correct option. Hence there is a discrepancy between the shown working and the marked option. Following the solution authority for MCQ answer extraction, the correct option is recorded as D.

Given: The question asks for the value of

$$80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta + \cos \theta}{9 + 16 \sin 2\theta} \, d\theta$$

Find: The matching option.

The source solution contains inconsistent working:

1. It changes the upper limit from $\frac{\pi}{4}$ to $\frac{\pi}{2}$.
2. It finally evaluates the integral as $4\ln 3$.
3. It simultaneously declares the correct option to be D.

Among the listed options,

• C is $4 \log 3$
• D is $3 \log 4$

So the displayed algebra supports C, while the solution's marks D. The extracted answer is therefore taken as D because the solution explicitly states "The Correct Option is D", but the inconsistency has been preserved here for transparency.