WBJEE 2024 Math Paper

To solve the given integral, let's consider the integral:$\int\limits_{0}^{\pi} e^{\cos^2 x} \cdot \cos^3((2n+1)x) \, dx$First, observe the integrand carefully. It includes the term $\cos^3((2n+1)x)$ which can be expanded using the trigonometric identity for cosine:$\cos^3(\theta) = \frac{1}{4} (\cos(3\theta) + 3 \cos(\theta))$ Applying this identity to the integrand, we have:$\cos^3((2n+1)x) = \frac{1}{4} (\cos(3(2n+1)x) + 3 \cos((2n+1)x))$Substitute this back into the integral:$\int\limits_{0}^{\pi} e^{\cos^2 x} \cdot \frac{1}{4} (\cos(3(2n+1)x) + 3 \cos((2n+1)x)) \, dx$This can be broken down into two separate integrals:$\frac{1}{4} \int\limits_{0}^{\pi} e^{\cos^2 x} \cos(3(2n+1)x) \, dx + \frac{3}{4} \int\limits_{0}^{\pi} e^{\cos^2 x} \cos((2n+1)x) \, dx$ Now, consider the properties of definite integrals and the cosine function over the interval from 0 to $\pi$. For odd multiples of $x$, the integral over the symmetric limits will cancel out to zero due to the periodicity and symmetry of the cosine function.Specifically, for any odd integer $k$ :$\int\limits_{0}^{\pi} \cos(kx) \, dx = 0$Therefore:$\int\limits_{0}^{\pi} e^{\cos^2 x} \cos(3(2n+1)x) \, dx = 0$$\int\limits_{0}^{\pi} e^{\cos^2 x} \cos((2n+1)x) \, dx = 0$Thus, combining these results:$\frac{1}{4} \times 0 + \frac{3}{4} \times 0 = 0$Therefore, the value of the integral is:0So, the correct option is:Option C: 0