If the integral $$525\int_0^{\pi/2} \sin(2x) \cos^{11/2}(x)(1 + \cos^{5/2}(x))^{1/2}dx$$ is equal to $$n\sqrt{2} - 64$$, then $$n$$ is equal to ________

We need to evaluate the integral $$525\int_0^{\pi/2} \sin(2x) \cos^{11/2}(x)(1 + \cos^{5/2}(x))^{1/2}\,dx$$ and find $$n$$ such that the result equals $$n\sqrt{2} - 64$$.

Since $$\sin(2x) = 2\sin x \cos x$$, the integral becomes $$I = 525 \int_0^{\pi/2} 2\sin x \cos x \cdot \cos^{11/2}(x) \cdot (1 + \cos^{5/2}(x))^{1/2}\,dx$$ which simplifies to $$I = 1050 \int_0^{\pi/2} \sin x \cdot \cos^{13/2}(x) \cdot (1 + \cos^{5/2}(x))^{1/2}\,dx$$.

Substituting $$t = \cos x$$ so that $$dt = -\sin x\,dx$$ and noting that when $$x = 0$$, $$t = 1$$ and when $$x = \pi/2$$, $$t = 0$$, the integral transforms to $$I = 1050 \int_1^0 t^{13/2}(1 + t^{5/2})^{1/2}(-dt)$$ which gives $$I = 1050 \int_0^1 t^{13/2}(1 + t^{5/2})^{1/2}\,dt$$.

Next, setting $$u = t^{5/2}$$ implies $$du = \tfrac{5}{2}t^{3/2}\,dt$$ so that $$dt = \tfrac{2}{5}t^{-3/2}\,du = \tfrac{2}{5}u^{-3/5}\,du$$, and noting that $$t^{13/2} = u^{13/5}$$. Substituting into the integral yields $$I = 1050 \int_0^1 u^{13/5}(1+u)^{1/2} \cdot \tfrac{2}{5}u^{-3/5}\,du = 420 \int_0^1 u^2(1+u)^{1/2}\,du$$.

To evaluate $$\int_0^1 u^2(1+u)^{1/2}\,du$$, let $$v = 1+u$$ so that $$u = v - 1$$ and $$du = dv$$, with the limits changing from $$u=0$$ to $$v=1$$ and from $$u=1$$ to $$v=2$$. This gives $$\int_1^2 (v-1)^2 v^{1/2}\,dv = \int_1^2 (v^2 - 2v + 1)v^{1/2}\,dv = \int_1^2 (v^{5/2} - 2v^{3/2} + v^{1/2})\,dv$$.

Integrating term by term yields $$\left[\frac{2v^{7/2}}{7} - \frac{4v^{5/2}}{5} + \frac{2v^{3/2}}{3}\right]_1^2$$. At $$v = 2$$ this expression is $$\frac{2 \cdot 2^{7/2}}{7} - \frac{4 \cdot 2^{5/2}}{5} + \frac{2 \cdot 2^{3/2}}{3} = \frac{16\sqrt{2}}{7} - \frac{16\sqrt{2}}{5} + \frac{4\sqrt{2}}{3}$$, and at $$v = 1$$ it equals $$\frac{2}{7} - \frac{4}{5} + \frac{2}{3} = \frac{30 - 84 + 70}{105} = \frac{16}{105}$$. Subtracting gives $$\sqrt{2}\left(\frac{16}{7} - \frac{16}{5} + \frac{4}{3}\right) - \frac{16}{105} = \sqrt{2}\cdot\frac{44}{105} - \frac{16}{105} = \frac{44\sqrt{2} - 16}{105}$$.

Hence the original integral is $$I = 420 \cdot \frac{44\sqrt{2} - 16}{105} = 4(44\sqrt{2} - 16) = 176\sqrt{2} - 64$$.

Comparing this with the form $$n\sqrt{2} - 64$$ shows that $$n = 176$$.

The value of $$n$$ is 176.