int_{frac{pi}{4}}^{frac{pi}{2}} cot ^9 x , dx | vedclass

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Question

(A) Let $I = \int_{\pi/4}^{\pi/2} \cot^9 x \, dx$. Substitute $t = \sin x$, so $dt = \cos x \, dx$. Since $\cot^9 x = \frac{\cos^9 x}{\sin^9 x} = \fra

Vedclass · Accepted Answer

$\frac{-7}{24} + \frac{1}{2} \log 2$

Vedclass · Answer

(A) Let $I = \int_{\pi/4}^{\pi/2} \cot^9 x \, dx$. Substitute $t = \sin x$, so $dt = \cos x \, dx$. Since $\cot^9 x = \frac{\cos^9 x}{\sin^9 x} = \frac{(\cos^2 x)^4 \cos x}{\sin^9 x} = \frac{(1-t^2)^4}{t^9}$, we have: $I = \int_{1/\sqrt{2}}^{1} \frac{(1-t^2)^4}{t^9} \, dt$. Expanding $(1-t^2)^4 = 1 - 4t^2 + 6t^4 - 4t^6 + t^8$: $I = \int_{1/\sqrt{2}}^{1} (t^{-9} - 4t^{-7} + 6t^{-5} - 4t^{-3} + t^{-1}) \, dt$. Integrating term by term: $I = \left[ \frac{t^{-8}}{-8} - 4\frac{t^{-6}}{-6} + 6\frac{t^{-4}}{-4} - 4\frac{t^{-2}}{-2} + \log|t| \right]_{1/\sqrt{2}}^{1}$. $I = \left[ -\frac{1}{8t^8} + \frac{2}{3t^6} - \frac{3}{2t^4} + \frac{2}{t^2} + \log t \right]_{1/\sqrt{2}}^{1}$. Evaluating at the limits: At $t=1$: $-\frac{1}{8} + \frac{2}{3} - \frac{3}{2} + 2 + \log 1 = \frac{-3 + 16 - 36 + 48}{24} = \frac{25}{24}$. At $t=1/\sqrt{2}$: $-\frac{1}{8(1/16)} + \frac{2}{3(1/8)} - \frac{3}{2(1/4)} + \frac{2}{1/2} + \log(1/\sqrt{2}) = -2 + \frac{16}{3} - 6 + 4 - \frac{1}{2} \log 2 = \frac{16}{3} - 4 - \frac{1}{2} \log 2 = \frac{4}{3} - \frac{1}{2} \log 2 = \frac{32}{24} - \frac{1}{2} \log 2$. $I = \frac{25}{24} - (\frac{32}{24} - \frac{1}{2} \log 2) = -\frac{7}{24} + \frac{1}{2} \log 2$.

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^9 x \, dx =$

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