_{π/4}^{π/2} ^9 x dx =

Correct Answer is : −724+12log⁡2-7/24 + 1/2 224−7+21log2

Correct Answer is : −724+12log⁡2-7/24 + 1/2 224−7+21log2

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TS EAMCET 04 May 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int\limits_{\pi/4}^{\pi/2} \cot^9

x dx =

Solution:

Consider the integral,

I \;\; =\int\limits_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cot^{9} x \, dx

\;\; =\int\limits_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \; \frac{\cos^{9} x}{\sin^{9} x} \, dx

Take

\sin x = t

then

\cos x \, dx = dt

I \;\; =\int\limits_{\;\frac{1}{\sqrt{2}}}^{1} \; \frac{(1-t^{2})^{4}}{t^{9}} \, dt

\;\; =\int\limits_{\;\frac{1}{\sqrt{2}}}^{1} \; \frac{1-4t^{2}+6t^{4}-4t^{6}+t^{8}}{t^{9}} \, dt

\;\; =\int\limits_{\;\frac{1}{\sqrt{2}}}^{1} \left[ t^{-9} - 4 t^{-7} + 6 t^{-5} - 4 t^{-3} + \;\frac{1}{t} \right] \, dt

\;\; =\left[ \;\frac{1}{8t^{8}} - \;\frac{4}{6t^{6}} + \;\frac{6}{4t^{4}} - \;\frac{4}{2t^{2}} + \log t \right]_{\;\frac{1}{\sqrt{2}}}^{1}

Then

I \;\; =\left[ \left( \;\frac{1}{8} - \;\frac{4}{6} + \;\frac{6}{4} - \;\frac{4}{2} + 0 \right) - \left( 2 - \;\frac{32}{6} - 6 - 4 - \;\frac{1}{2} \log 2 \right) \right]

\;\; = \;\frac{-7}{24} + \;\frac{1}{2} \log 2

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