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Question Number 39383 by maxmathsup by imad last updated on 05/Jul/18

calculate  ∫_0 ^(π/3)       ((sinxdx)/(cosx(2+ln(cosx))) .

$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\:\:\frac{{sinxdx}}{{cosx}\left(\mathrm{2}+{ln}\left({cosx}\right)\right.}\:. \\ $$

Commented by math khazana by abdo last updated on 06/Jul/18

chanvement  cosx =t give  I  = ∫_1 ^(1/2)      ((√(1−t^2 ))/(t(2+ln(t)))) ((−dt)/(√(1−t^2 )))  = ∫_(1/2) ^1       (dt/(t(2 +ln(t)))) then we do the chang.  ln(t)=u ⇒I = ∫_(−ln(2)) ^0  ((e^u  du)/(e^u (2 +u)))  = ∫_(−ln(2)) ^0     (du/(2+u)) =[ln∣2+u∣]_(−ln(2)) ^0   =ln(2) −ln(2−ln(2)) .

$${chanvement}\:\:{cosx}\:={t}\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\:\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}\left(\mathrm{2}+{ln}\left({t}\right)\right)}\:\frac{−{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{dt}}{{t}\left(\mathrm{2}\:+{ln}\left({t}\right)\right)}\:{then}\:{we}\:{do}\:{the}\:{chang}. \\ $$$${ln}\left({t}\right)={u}\:\Rightarrow{I}\:=\:\int_{−{ln}\left(\mathrm{2}\right)} ^{\mathrm{0}} \:\frac{{e}^{{u}} \:{du}}{{e}^{{u}} \left(\mathrm{2}\:+{u}\right)} \\ $$$$=\:\int_{−{ln}\left(\mathrm{2}\right)} ^{\mathrm{0}} \:\:\:\:\frac{{du}}{\mathrm{2}+{u}}\:=\left[{ln}\mid\mathrm{2}+{u}\mid\right]_{−{ln}\left(\mathrm{2}\right)} ^{\mathrm{0}} \\ $$$$={ln}\left(\mathrm{2}\right)\:−{ln}\left(\mathrm{2}−{ln}\left(\mathrm{2}\right)\right)\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18

t=2+lncosx  t=2   when x=0  t=2+ln((1/2))=2−ln2  dt=((−sinx)/(cosx))dx  ∫_2 ^(2−ln2)    ((−dt)/t)  −∣lnt∣_2 ^(2−ln2)   −ln∣((2−ln2)/2)∣

$${t}=\mathrm{2}+{lncosx} \\ $$$${t}=\mathrm{2}\:\:\:{when}\:{x}=\mathrm{0} \\ $$$${t}=\mathrm{2}+{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}−{ln}\mathrm{2} \\ $$$${dt}=\frac{−{sinx}}{{cosx}}{dx} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{2}−{ln}\mathrm{2}} \:\:\:\frac{−{dt}}{{t}} \\ $$$$−\mid{lnt}\mid_{\mathrm{2}} ^{\mathrm{2}−{ln}\mathrm{2}} \\ $$$$−{ln}\mid\frac{\mathrm{2}−{ln}\mathrm{2}}{\mathrm{2}}\mid \\ $$

Commented by math khazana by abdo last updated on 06/Jul/18

answer correct sir Tanmay thanks.

$${answer}\:{correct}\:{sir}\:{Tanmay}\:{thanks}. \\ $$

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