Question Number 35588 by abdo mathsup 649 cc last updated on 20/May/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{sinx}\:{cos}\left({cosx}\right)}{\mathrm{1}+\mathrm{2}{sin}\left({cosx}\right)}{dx} \\ $$
Commented by abdo imad last updated on 21/May/18
![we remark that (sin(cosx))^′ =−sinx cos(cosx) ⇒ I =−(1/2) ∫_0 ^(π/3) (((1+2sin(cosx))^′ )/(1+2 sin(cosx)))dx =−(1/2)[ln∣ 1+2sin(cosx)∣]_0 ^(π/3) = =−(1/2){ln(1+2sin((1/2)) −ln(1 +2sin(1)}](Q35653.png)
$${we}\:{remark}\:{that}\:\left({sin}\left({cosx}\right)\right)^{'} \:=−{sinx}\:{cos}\left({cosx}\right)\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\left(\mathrm{1}+\mathrm{2}{sin}\left({cosx}\right)\right)^{'} }{\mathrm{1}+\mathrm{2}\:{sin}\left({cosx}\right)}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\:\mathrm{1}+\mathrm{2}{sin}\left({cosx}\right)\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} = \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{1}+\mathrm{2}{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:−{ln}\left(\mathrm{1}\:+\mathrm{2}{sin}\left(\mathrm{1}\right)\right\}\right.\right. \\ $$
Answered by RAMANUJAN last updated on 20/May/18
![[ln{(2Π+3)/3}]/2−0 is the correct answer](Q35598.png)
$$\left[\mathrm{ln}\left\{\left(\mathrm{2}\Pi+\mathrm{3}\right)/\mathrm{3}\right\}\right]/\mathrm{2}−\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer} \\ $$