approximation formula for the integral Get an approximation formula for the following integral:
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$

$$\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
First observe the powers of $cos$ and $sin$
$$ cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) = cos^{2n-2k}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) =
   (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y)$$
$$
\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy 

$$
If we use the substitution $t = sin(y) \Rightarrow dt = cos(y) \hspace{4pt} dy$
$t = 0$ when $y=0$ and $t=1$ when $y=\frac{\pi}{2}$
$$
\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^2)^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy 
$$
$$
= \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt
$$
The expression
$$
\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
=  \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right)
$$
$$
= \left( \frac{1}{35} \right)^{k-1}\int_0^1 \left( \sum_{k=1}^{n} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt
$$
Notice that the sum is a geometric series with first term $\frac{1}{35}(1-t^{2})^{n}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})^{2}}$
The expression therefore simplifies to
$$
= \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^2)^{(n-k)} \right) dt
$$
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
= \left( \frac{1}{35} \right)^{k-1} \sum_{k=1}^n \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right)
$$
$$
= \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt
$$
Notice that the sum is a geometric series with first term $(1-t^{2})^{(n-1)}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})}$
The expression therefore simplifies to
$$
= \int_0^1 \left( \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( t^{(2k-2)} (1-t^{2})^{(n-k)} \right) \right) dt
$$
This expression turns out to be the summation of constant multiple of Beta function
