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Step-by-Step Solution
Step 1: Express sin(2x) in Terms of sin(x) and cos(x)
We need to evaluate the integral
525 \int_{0}^{\frac{\pi}{2}} \sin 2x \,\cos^{\frac{11}{2}} x \Bigl(1 + \bigl(\cos x\bigr)^{\frac{5}{2}}\Bigr)^{\frac{1}{2}} \, dx ,
which is given to be equal to (n\sqrt{2} - 64) . We want to find the value of n .
First, recall the identity \sin(2x) = 2\,\sin(x)\,\cos(x) . Let us denote the integral (without the leading factor 525) as I :
I = \int_{0}^{\frac{\pi}{2}} \sin(2x)\,\cos^{\frac{11}{2}}(x)\,\sqrt{1 + \cos^{\frac{5}{2}}(x)} \, dx .
Substitute \sin(2x) = 2\sin(x)\cos(x) :
I = \int_{0}^{\frac{\pi}{2}} 2\,\sin(x)\,\cos(x)\,\cos^{\tfrac{11}{2}}(x)\, \sqrt{1 + \cos^{\tfrac{5}{2}}(x)} \, dx .
Combine the powers of \cos(x) :
\cos(x)\,\cos^{\frac{11}{2}}(x) = \cos^{\frac{13}{2}}(x) .
Thus, the integral simplifies to:
I = 2 \int_{0}^{\frac{\pi}{2}} \sin(x)\,\cos^{\frac{13}{2}}(x)\, \sqrt{1 + \cos^{\tfrac{5}{2}}(x)} \, dx .
Step 2: Make the Substitution t^2 = \cos(x)
Let us set t^2 = \cos(x) . Then differentiating both sides gives:
2t \, dt = -\sin(x)\,dx \quad \Longrightarrow \quad \sin(x)\,dx = -2t\,dt .
The limits of integration change accordingly:
When x = 0 , \cos(0) = 1 \Rightarrow t^2 = 1 \Rightarrow t = 1.
When x = \frac{\pi}{2} , \cos\!\bigl(\tfrac{\pi}{2}\bigr) = 0 \Rightarrow t^2 = 0 \Rightarrow t = 0.
Hence, I becomes:
I = 2 \int_{0}^{\tfrac{\pi}{2}} \sin(x)\,\cos^{\tfrac{13}{2}}(x)\,\sqrt{1 + \cos^{\tfrac{5}{2}}(x)} \, dx = 2 \int_{1}^{0} (t^2)^{\tfrac{13}{2}}\,\sqrt{1 + (t^2)^{\tfrac{5}{2}}}\,(-2t)\,dt .
Since (t^2)^{\tfrac{13}{2}} = t^{13} and (t^2)^{\tfrac{5}{2}} = t^5 , this becomes:
I = -4 \int_{1}^{0} t^{13}\,\sqrt{1 + t^5}\,t \, dt = -4 \int_{1}^{0} t^{14}\,\sqrt{1 + t^5}\, dt .
Reversing the limits from 0 to 1 introduces a minus sign, which cancels the existing negative sign:
I = 4 \int_{0}^{1} t^{14}\,\sqrt{1 + t^5}\, dt .
Step 3: Substitute 1 + t^5 = k^2
Set 1 + t^5 = k^2 . Then t^5 = k^2 - 1 . Differentiate both sides:
5\,t^4\,dt = 2k\,dk \quad \Longrightarrow \quad t^4\,dt = \frac{2k}{5}\,dk .
Also, observe:
t^{14} = t^{10} \cdot t^4 = (t^5)^2 \cdot t^4 = (k^2 - 1)^2 \, t^4, \quad \sqrt{1 + t^5} = k .
Hence, t^{14}\,\sqrt{1 + t^5}\,dt = (k^2 - 1)^2 \, k \, \bigl(t^4\,dt\bigr) = (k^2 - 1)^2 \, k \, \frac{2k}{5}\,dk = \frac{2k^2}{5}\,(k^2 - 1)^2\,dk .
The new limits for k come from:
When t = 0 , k^2 = 1 \Rightarrow k = 1.
When t = 1 , k^2 = 2 \Rightarrow k = \sqrt{2}.
Thus,
I = 4 \int_{0}^{1} t^{14}\,\sqrt{1 + t^5}\,dt = 4 \int_{1}^{\sqrt{2}} \frac{2k^2}{5}\,(k^2 - 1)^2\,dk = \frac{8}{5}\,\int_{1}^{\sqrt{2}} k^2\,(k^2 - 1)^2\,dk .
Next, expand (k^2 - 1)^2 = k^4 - 2k^2 + 1 , so
k^2 (k^4 - 2k^2 + 1) = k^6 - 2k^4 + k^2 .
Hence,
I = \frac{8}{5} \int_{1}^{\sqrt{2}} \bigl(k^6 - 2k^4 + k^2\bigr)\,dk .
Step 4: Integrate and Evaluate
We integrate each term:
\int k^6\, dk = \frac{k^7}{7}, \quad \int (-2k^4)\, dk = -2 \cdot \frac{k^5}{5}, \quad \int k^2\, dk = \frac{k^3}{3} .
Thus,
\int_{1}^{\sqrt{2}} \bigl(k^6 - 2k^4 + k^2\bigr)\, dk = \left[\frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3}\right]_{1}^{\sqrt{2}} .
So,
I = \frac{8}{5}\,\Biggl(\,\Bigl[\frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3}\Bigr]_{1}^{\sqrt{2}}\Biggr) .
We then substitute k = \sqrt{2} and subtract the expression at k=1 . That yields a numeric expression for I .
Step 5: Multiply by 525 and Find n
The original integral, multiplied by 525, is given to be n \sqrt{2} - 64 . In other words,
525 \, I = n \sqrt{2} - 64 .
Careful algebraic evaluation of I shows that
525 \, I = 176\,\sqrt{2} \;-\; 64 .
Hence, by direct comparison, n = 176.
Final Answer: 176
