MCQ
$\int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} = $
- A$\log \left( {\frac{8}{9}} \right)$
- ✓$\log \left( {\frac{9}{8}} \right)$
- C$\log (8 \times 9)$
- DNone of these
We put $\cos x = t \Rightarrow - \sin x\,dx = dt,$ then
$I = \int_0^1 {\frac{{t.dt}}{{{t^2} + 3t + 2}} = \int_0^1 {\left[ {\frac{2}{{t + 2}} - \frac{1}{{t + 1}}} \right]} } \,dt$
$ = [2\log (t + 2) - \log (\,t + 1)]_0^1$
$ = [2\log 3 - \log 2 - 2\log 2]$
$ = [2\log 3 - 3\log 2] = [\log 9 - \log 8] $
$= \log \left( {\frac{9}{8}} \right)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{1}{3\ln\text{x}}$
$\frac{1}{3\ln\text{x}}-\frac{1}{2\ln\text{x}}$
$\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$
$\frac{3\text{x}^2}{\ln\text{x}}$