Question

$\int_{0}^{\pi/2} \frac{\sin x \cos x dx}{\cos^2 x+3 \cos x+2} =$

Answer 1

$\ln(9/8)$

Answer 2

Let the integral be $I$. $I = \int_{0}^{\pi/2} \frac{\sin x \cos x dx}{\cos^2 x+3 \cos x+2}$ Let $u = \cos x$. Then $du = -\sin x dx$. When $x=0$, $u = \cos(0) = 1$. When $x=\pi/2$, $u = \cos(\pi/2) = 0$. Substituting these into the integral, we get: $I = \int_{1}^{0} \frac{u (-du)}{u^2+3u+2}$ $I = \int_{1}^{0} \frac{-u du}{u^2+3u+2}$ We can swap the limits of integration by changing the sign of the integral: $I = \int_{0}^{1} \frac{u du}{u^2+3u+2}$ Factor the denominator: $u^2+3u+2 = (u+1)(u+2)$. So the integral becomes: $I = \int_{0}^{1} \frac{u}{(u+1)(u+2)} du$ We use partial fraction decomposition for the integrand: $\frac{u}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2}$ Multiplying both sides by $(u+1)(u+2)$, we get: $u = A(u+2) + B(u+1)$ To find $A$, set $u=-1$: $-1 = A(-1+2) + B(-1+1) \implies -1 = A(1) + B(0) \implies A = -1$ To find $B$, set $u=-2$: $-2 = A(-2+2) + B(-2+1) \implies -2 = A(0) + B(-1) \implies -2 = -B \implies B = 2$ So, the integrand is: $\frac{u}{(u+1)(u+2)} = \frac{-1}{u+1} + \frac{2}{u+2}$ Now, we integrate the decomposed fractions: $I = \int_{0}^{1} \left( \frac{2}{u+2} - \frac{1}{u+1} \right) du$ $I = \left[ 2 \ln|u+2| - \ln|u+1| \right]_{0}^{1}$ Evaluate the definite integral at the limits: $I = (2 \ln|1+2| - \ln|1+1|) - (2 \ln|0+2| - \ln|0+1|)$ $I = (2 \ln 3 - \ln 2) - (2 \ln 2 - \ln 1)$ Since $\ln 1 = 0$: $I = (2 \ln 3 - \ln 2) - (2 \ln 2 - 0)$ $I = 2 \ln 3 - \ln 2 - 2 \ln 2$ $I = 2 \ln 3 - 3 \ln 2$ Using the logarithm property $a \ln b = \ln(b^a)$: $I = \ln(3^2) - \ln(2^3)$ $I = \ln 9 - \ln 8$ Using the logarithm property $\ln a - \ln b = \ln(a/b)$: $I = \ln \left(\frac{9}{8}\right)$