a Let k ∈ N. Find an expression in terms of k for ∫ π20 sin2k x dx

The integral of sin^2k(x) from 0 to π/2 can be solved using the power-reduction identity.

The power-reduction identity is given by:

sin^2(x) = 1/2 - 1/2 cos(2x)

So, sin^2k(x) = (1/2 - 1/2 cos(2x))^k

Now, we can integrate this from 0 to π/2:

∫ from 0 to π/2 sin^2k(x) dx = ∫ from 0 to π/2 (1/2 - 1/2 cos(2x))^k dx

This integral can be solved using the binomial theorem, which states that (a - b)^n = Σ (n choose k) a^(n-k) * b^k from k=0 to n.

Applying the binomial theorem to the integral gives:

∫ from 0 to π/2 sin^2k(x) dx = ∫ from 0 to π/2 Σ (k choose j) (1/2)^(k-j) * (-1/2 cos(2x))^j dx from j=0 to k

This integral can be solved term by term. The integral of cos(2x) from 0 to π/2 is 0, so any term with j > 0 will integrate to 0. The only term that does not integrate to 0 is the term with j = 0.

So, ∫ from 0 to π/2 sin^2k(x) dx = (k choose 0) * (1/2)^k * ∫ from 0 to π/2 dx = (1/2)^k * [x] from 0 to π/2 = (1/2)^k * π/2

So, the integral of sin^2k(x) from 0 to π/2 in terms of k is (1/2)^k * π/2.

The integral of sin^2k(x) from 0 to π/2 can be solved using the power-reduction identity.

The power-reduction identity is given by:

sin^2(x) = 1/2 - 1/2 cos(2x)

So, sin^2k(x) = (1/2 - 1/2 cos(2x))^k

Now, we can integrate this from 0 to π/2:

∫ from 0 to π/2 sin^2k(x) dx = ∫ from 0 to π/2 (1/2 - 1/2 cos(2x))^k dx

This integral can be solved using the binomial theorem, which states that (a - b)^n = Σ (n choose i) a^(n-i) * b^i from i=0 to n.

Applying the binomial theorem to the integral gives:

∫ from 0 to π/2 sin^2k(x) dx = Σ from i=0 to k (k choose i) * ∫ from 0 to π/2 (1/2)^(k-i) * (-1/2 cos(2x))^i dx

The integral of cos(2x) from 0 to π/2 is 0, so the only term that contributes to the sum is the i=0 term:

∫ from 0 to π/2 sin^2k(x) dx = (k choose 0) * ∫ from 0 to π/2 (1/2)^k dx = (1/2)^k * [x] from 0 to π/2 = (1/2)^k * π/2

So, the integral of sin^2k(x) from 0 to π/2 is (1/2)^k * π/2.