15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T65b,

To solve the given integral, let's consider the integral:

$$\int_{0}^{\pi} e^{\cos^2 x} \cdot \cos^3((2n+1)x) \, dx$$

First, observe the integrand carefully. It includes the term $$\cos^3((2n+1)x)$$ which can be expanded using the trigonometric identity for cosine:

$$\cos^3(\theta) = \frac{1}{4} \left( \cos(3\theta) + 3\cos(\theta) \right)$$

Applying this identity to the integrand, we have:

$$\cos^3((2n+1)x) = \frac{1}{4} \left( \cos(3(2n+1)x) + 3\cos((2n+1)x) \right)$$

Substitute this back into the integral:

$$\int_{0}^{\pi} e^{\cos^2 x} \cdot \frac{1}{4} \left( \cos(3(2n+1)x) + 3\cos((2n+1)x) \right) \, dx$$

This can be broken down into two separate integrals:

$$\frac{1}{4} \int_{0}^{\pi} e^{\cos^2 x} \cos(3(2n+1)x) \, dx + \frac{3}{4} \int_{0}^{\pi} e^{\cos^2 x} \cos((2n+1)x) \, dx$$

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 $$k$$:

$$\int_{0}^{\pi} \cos(kx) \, dx = 0$$

Therefore:

$$\int_{0}^{\pi} e^{\cos^2 x} \cos(3(2n+1)x) \, dx = 0$$

and

$$\int_{0}^{\pi} e^{\cos^2 x} \cos((2n+1)x) \, dx = 0$$

Thus, combining these results:

$$\frac{1}{4} \times 0 + \frac{3}{4} \times 0 = 0$$

Therefore, the value of the integral is:

Definite Integration

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