Evaluate the following integrals: ^ 2 _ 0 ^2x+3 +2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$

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Answer

Let $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$ Then,
Let $\cos\text{x}=\text{t},$ Then, $-\sin\text{x dx}=\text{dt}$
When, $\text{x}=0,\text{ t}=1$ and $\text{x}=\frac{\pi}{2},\text{ t}=0$
$\therefore\ \text{I}=-\int^\limits0_1\frac{\text{t dt}}{\text{t}^2+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\frac{-\text{t dt}}{\text{t}^3+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\Big(\frac{1}{(\text{t}+1)}-\frac{2}{(\text{t}+2)}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\log(\text{t}+1)-2\log(\text{t}+2)\Big]^0_1$
$\Rightarrow\text{I}=\bigg[\log\frac{(\text{t}+1)}{(\text{t}+2)^2}\bigg]^1_0$
$\Rightarrow\text{I}=\bigg[\log\Big(\frac{1}{4}\Big)-\log\Big(\frac{2}{9}\Big)\bigg]^1_0$
$\Rightarrow\text{I}=\log\frac{9}{8}$

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