WBJEE 2024 Math Paper

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

π

∫

0

ecos2x⋅cos3((2n+1)x)‌dx
First, observe the integrand carefully. It includes the term cos3((2n+1)x) which can be expanded using the trigonometric identity for cosine:
cos3(θ)=‌

1

4

(cos(3θ)+3‌cos(θ))

Applying this identity to the integrand, we have:
cos3((2n+1)x)=‌

1

4

(cos(3(2n+1)x)+3‌cos((2n+1)x))
Substitute this back into the integral:

π

∫

0

ecos2x⋅‌

1

4

(cos(3(2n+1)x)+3‌cos((2n+1)x))‌dx
This can be broken down into two separate integrals:
‌

1

4

‌

π

∫

0

ecos2x‌cos(3(2n+1)x)‌dx+‌

3

4

‌

π

∫

0

ecos2x‌cos((2n+1)x)‌dx

Now, consider the properties of definite integrals and the cosine function over the interval from 0 to π. 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 :

π

∫

0

cos(kx)‌dx=0
Therefore:

π

∫

0

ecos2x‌cos(3(2n+1)x)‌dx=0

π

∫

0

ecos2x‌cos((2n+1)x)‌dx=0
Thus, combining these results:
‌

1

4

×0+‌

3

4

×0=0
Therefore, the value of the integral is:
0
So, the correct option is:
Option C: 0