Key Concepts and Formulas

• Definite Integral by Substitution: If $I = \int_{a}^{b} f(g(x)) g'(x) dx$, we can substitute $u = g(x)$, so $du = g'(x) dx$. The limits of integration change from $x=a$ to $u=g(a)$ and from $x=b$ to $u=g(b)$. The integral becomes $I = \int_{g(a)}^{g(b)} f(u) du$.
• Trigonometric Identity: $\sin 2x = 2 \sin x \cos x$.
• Power Rule for Integration: $\int u^n du = \frac{u^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Let the given integral be $I$. I = 525 \int_\limits0^{\frac{\pi}{2}} \sin 2 x \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x

Step 1: Simplify the integrand using the double angle identity. We use the identity $\sin 2x = 2 \sin x \cos x$ to rewrite the integral. I = 525 \int_\limits0^{\frac{\pi}{2}} (2 \sin x \cos x) \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x I = 525 \times 2 \int_\limits0^{\frac{\pi}{2}} \sin x \cos x \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x I = 1050 \int_\limits0^{\frac{\pi}{2}} \sin x \cos ^{\frac{11}{2} + 1} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x I = 1050 \int_\limits0^{\frac{\pi}{2}} \sin x \cos ^{\frac{13}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x

Step 2: Apply substitution to simplify the integral. Observe that the derivative of $\cos x$ is $-\sin x$, which is present in the integrand. Also, the term $(1+\operatorname{Cos}^{\frac{5}{2}} x)$ inside the square root suggests a substitution. Let $u = \cos x$. Then, $du = -\sin x \, dx$.

Now, we need to change the limits of integration: When $x = 0$, $u = \cos 0 = 1$. When $x = \frac{\pi}{2}$, $u = \cos \frac{\pi}{2} = 0$.

Substituting $u = \cos x$ and $du = -\sin x \, dx$ into the integral, we get: I = 1050 \int_\limits1^{0} u^{\frac{13}{2}} (1+u^{\frac{5}{2}})^{\frac{1}{2}} (-du) I = -1050 \int_\limits1^{0} u^{\frac{13}{2}} (1+u^{\frac{5}{2}})^{\frac{1}{2}} du

Step 3: Reverse the limits of integration. We can reverse the limits of integration by changing the sign of the integral. I = 1050 \int_\limits0^{1} u^{\frac{13}{2}} (1+u^{\frac{5}{2}})^{\frac{1}{2}} du

Step 4: Apply another substitution. The term $(1+u^{\frac{5}{2}})^{\frac{1}{2}}$ suggests a further substitution. Let $v = 1+u^{\frac{5}{2}}$. Then, $dv = \frac{5}{2} u^{\frac{5}{2}-1} du = \frac{5}{2} u^{\frac{3}{2}} du$.

However, we have $u^{\frac{13}{2}} du$ in the integral, not $u^{\frac{3}{2}} du$. We can rewrite $u^{\frac{13}{2}}$ as $u^{10/2} \cdot u^{3/2} = u^5 \cdot u^{3/2}$. This still doesn't seem to directly fit.

Let's try a different substitution for the entire term inside the square root. Let $v^2 = 1+u^{\frac{5}{2}}$. This might lead to complex derivatives.

Let's re-examine the integral: I = 1050 \int_\limits0^{1} u^{\frac{13}{2}} (1+u^{\frac{5}{2}})^{\frac{1}{2}} du. Consider the term $u^{\frac{13}{2}}$. We can write $u^{\frac{13}{2}} = u^{\frac{10}{2}} \cdot u^{\frac{3}{2}} = u^5 \cdot u^{\frac{3}{2}}$. So, I = 1050 \int_\limits0^{1} u^5 \cdot u^{\frac{3}{2}} (1+u^{\frac{5}{2}})^{\frac{1}{2}} du.

Now, let's use the substitution $v = 1+u^{\frac{5}{2}}$. Then $dv = \frac{5}{2} u^{\frac{3}{2}} du$. This means $u^{\frac{3}{2}} du = \frac{2}{5} dv$.

From $v = 1+u^{\frac{5}{2}}$, we have $u^{\frac{5}{2}} = v-1$. We also need to express $u^5$ in terms of $v$. $u^{\frac{5}{2}} = v-1 \implies (u^{\frac{5}{2}})^2 = (v-1)^2 \implies u^5 = (v-1)^2$.

Now, let's change the limits of integration for $v$: When $u = 0$, $v = 1+0^{\frac{5}{2}} = 1$. When $u = 1$, $v = 1+1^{\frac{5}{2}} = 1+1 = 2$.

Substitute these into the integral: I = 1050 \int_\limits1^{2} (v-1)^2 \cdot (v)^{\frac{1}{2}} \cdot \frac{2}{5} dv I = 1050 \times \frac{2}{5} \int_\limits1^{2} (v^2 - 2v + 1) v^{\frac{1}{2}} dv I = 420 \int_\limits1^{2} (v^{\frac{5}{2}} - 2v^{\frac{3}{2}} + v^{\frac{1}{2}}) dv

Step 5: Integrate with respect to v. Now we use the power rule for integration: $\int v^n dv = \frac{v^{n+1}}{n+1}$. $I = 420 \left[ \frac{v^{\frac{5}{2}+1}}{\frac{5}{2}+1} - 2 \frac{v^{\frac{3}{2}+1}}{\frac{3}{2}+1} + \frac{v^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_1^2$ $I = 420 \left[ \frac{v^{\frac{7}{2}}}{\frac{7}{2}} - 2 \frac{v^{\frac{5}{2}}}{\frac{5}{2}} + \frac{v^{\frac{3}{2}}}{\frac{3}{2}} \right]_1^2$ $I = 420 \left[ \frac{2}{7} v^{\frac{7}{2}} - \frac{4}{5} v^{\frac{5}{2}} + \frac{2}{3} v^{\frac{3}{2}} \right]_1^2$

Step 6: Evaluate the definite integral. Substitute the upper and lower limits: $I = 420 \left[ \left( \frac{2}{7} (2)^{\frac{7}{2}} - \frac{4}{5} (2)^{\frac{5}{2}} + \frac{2}{3} (2)^{\frac{3}{2}} \right) - \left( \frac{2}{7} (1)^{\frac{7}{2}} - \frac{4}{5} (1)^{\frac{5}{2}} + \frac{2}{3} (1)^{\frac{3}{2}} \right) \right]$

Let's simplify the terms involving powers of 2: $2^{\frac{7}{2}} = 2^3 \cdot 2^{\frac{1}{2}} = 8\sqrt{2}$ $2^{\frac{5}{2}} = 2^2 \cdot 2^{\frac{1}{2}} = 4\sqrt{2}$ $2^{\frac{3}{2}} = 2^1 \cdot 2^{\frac{1}{2}} = 2\sqrt{2}$

The terms involving powers of 1 are just 1.

So, the expression becomes: $I = 420 \left[ \left( \frac{2}{7} (8\sqrt{2}) - \frac{4}{5} (4\sqrt{2}) + \frac{2}{3} (2\sqrt{2}) \right) - \left( \frac{2}{7} - \frac{4}{5} + \frac{2}{3} \right) \right]$ $I = 420 \left[ \left( \frac{16\sqrt{2}}{7} - \frac{16\sqrt{2}}{5} + \frac{4\sqrt{2}}{3} \right) - \left( \frac{2}{7} - \frac{4}{5} + \frac{2}{3} \right) \right]$

Combine the terms with $\sqrt{2}$: $\frac{16}{7} - \frac{16}{5} + \frac{4}{3} = \frac{16 \times 15 - 16 \times 21 + 4 \times 35}{105} = \frac{240 - 336 + 140}{105} = \frac{44}{105}$ So, the first part is $\frac{44\sqrt{2}}{105}$.

Combine the constant terms: $\frac{2}{7} - \frac{4}{5} + \frac{2}{3} = \frac{2 \times 15 - 4 \times 21 + 2 \times 35}{105} = \frac{30 - 84 + 70}{105} = \frac{16}{105}$

Now substitute these back into the expression for $I$: $I = 420 \left[ \frac{44\sqrt{2}}{105} - \frac{16}{105} \right]$ $I = \frac{420}{105} (44\sqrt{2} - 16)$ Since $420 = 4 \times 105$, we have: $I = 4 (44\sqrt{2} - 16)$ $I = 176\sqrt{2} - 64$

Step 7: Equate with the given expression and find n. We are given that the integral is equal to $(n \sqrt{2}-64)$. So, $176\sqrt{2} - 64 = n \sqrt{2} - 64$.

Comparing the coefficients of $\sqrt{2}$ and the constant terms, we get: $n = 176$.

Common Mistakes & Tips

• Sign errors during substitution: Be very careful when changing the differential ($dx$ to $du$) and reversing the limits of integration. A common mistake is forgetting the negative sign when reversing limits.
• Algebraic errors with fractional exponents: Simplify terms like $u^{\frac{13}{2}}$ and $u^{\frac{5}{2}}$ carefully, especially when making substitutions. Writing $u^{\frac{13}{2}} = u^5 \cdot u^{\frac{3}{2}}$ is key.
• Calculation errors in evaluating definite integrals: Double-check the arithmetic when substituting the upper and lower limits and combining terms.

Summary

The problem involves evaluating a definite integral with trigonometric functions and fractional exponents. The solution proceeds by first simplifying the integrand using the double angle identity for sine. Then, a series of substitutions are employed to transform the integral into a polynomial in a new variable, which can be integrated using the power rule. Finally, the limits of integration are applied, and the result is compared with the given expression to find the value of $n$. The key steps involved were trigonometric simplification, two successive substitutions, and careful evaluation of the resulting definite integral.

The final answer is \boxed{176}.