int_0^{pi /2} {frac{{sin xcos x,dx}}{{{{cos } | vedclass

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(B) Let $I = \int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} $. Substitute $\cos x = t$,then $-\sin x\,dx = dt$. When $x = 0

Vedclass · Accepted Answer

$\log \left( {\frac{9}{8}} \right)$

Vedclass · Answer

(B) Let $I = \int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} $. Substitute $\cos x = t$,then $-\sin x\,dx = dt$. When $x = 0$,$t = 1$. When $x = \pi/2$,$t = 0$. Thus,$I = \int_1^0 {\frac{-t\,dt}{{{t^2} + 3t + 2}}} = \int_0^1 {\frac{t\,dt}{{(t+1)(t+2)}}} $. Using partial fractions,$\frac{t}{(t+1)(t+2)} = \frac{2}{t+2} - \frac{1}{t+1}$. $I = \int_0^1 {\left( {\frac{2}{{t + 2}} - \frac{1}{{t + 1}}} \right)} \,dt$. $I = [2\log |t + 2| - \log |t + 1|]_0^1$. $I = (2\log 3 - \log 2) - (2\log 2 - \log 1)$. Since $\log 1 = 0$,$I = 2\log 3 - 3\log 2 = \log 3^2 - \log 2^3 = \log 9 - \log 8 = \log \left( {\frac{9}{8}} \right)$.

$\int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} = $

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