=∫0π2 dx4+5cos⁡x=∫01 2dt1+t24+51−t21+t2 put tan⁡x2=tcos⁡x=1−t21+t2,dx=2dt1+t2 LL_UL_t=tan⁡0t=tan⁡π4t=0t=1=∫01 2dt4+4t2+5−5t2=2∫01 dt9−t2=212(3)log⁡3+t3−t01=13log⁡3+t3−t01=13log⁡42−log⁡(1)==13log⁡2

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$\int_{0}^{\frac{π}{2}} \frac{d x}{4 + 5 \cos x} =$

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$\frac{1}{2} \log 2$

$\frac{1}{3} \log 3$

$\frac{1}{5} \log 2$

$\frac{1}{3} \log 2$

answer is D.

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Detailed Solution

$= \int_{0}^{\frac{π}{2}} \frac{d x}{4 + 5 \cos x} = \int_{0}^{1} \frac{\frac{2 d t}{1 + t^{2}}}{4 + 5 (\frac{1 - t^{2}}{1 + t^{2}})} put \tan \frac{x}{2} = t \cos x = \frac{1 - t^{2}}{1 + t^{2}}, d x = \frac{2 d t}{1 + t^{2}}$

$\begin{array}{lc} \underline{LL} & \underline{UL} \\ t = \tan 0 & t = \tan \frac{π}{4} \\ t = 0 & t = 1 \end{array}$

$\begin{array}{l} = \int_{0}^{1} \frac{2 d t}{4 + 4 t^{2} + 5 - 5 t^{2}} \\ = 2 \int_{0}^{1} \frac{d t}{9 - t^{2}} = 2 {[\frac{1}{2 (3)} \log (\frac{3 + t}{3 - t})]}_{0}^{1} \\ = \frac{1}{3} \log {|\frac{3 + t}{3 - t}|}_{0}^{1} = \frac{1}{3} [\log (\frac{4}{2}) - \log (1)] == \frac{1}{3} \log 2 \end{array}$