Question Number 17033 by arnabpapu550@gmail.com last updated on 30/Jun/17
$$\int_{\frac{\Pi\:}{\mathrm{2}}} ^{\:\mathrm{0}} \:\frac{\mathrm{sinx}\:\mathrm{cosx}\:\mathrm{dx}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{x}+\mathrm{3sin}^{\mathrm{2}} \mathrm{x}} \\ $$
Answered by sma3l2996 last updated on 30/Jun/17
![t=sinx⇒dt=cosxdx I=∫_1 ^0 ((tdt)/(2+t^2 ))=(1/2)[ln∣2+t^2 ∣]_1 ^0 =(1/2)(ln2−ln3)=(1/2)ln(2/3)](https://www.tinkutara.com/question/Q17038.png)
$${t}={sinx}\Rightarrow{dt}={cosxdx} \\ $$$${I}=\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{tdt}}{\mathrm{2}+{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\mathrm{2}+{t}^{\mathrm{2}} \mid\right]_{\mathrm{1}} ^{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\mathrm{2}−{ln}\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by arnabpapu550@gmail.com last updated on 30/Jun/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$