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Question Number 108748 by john santu last updated on 18/Aug/20
Answered by bemath last updated on 19/Aug/20
by parts { ((u=ln (sin x) ⇒du=((cos x)/(sin x)) dx)),((v=∫ sin x dx=−cos x)) :} J=−cos x ln (sin x)+∫ ((cos^2 x dx)/(sin x)) J=−cos x ln (sin x)+∫ ((1−sin^2 x)/(sin x)) dx J=−cos x ln (sin x)+∫ cosec x dx −∫sin xdx J=−cos xln (sin x)+cos x+ln ∣cosec x−cot x∣ J={cos x (1−ln (sin x))+ln ∣((1−cos x)/(sin x))∣}_0 ^(π/2) J=lim_(b→0) {cos x(1−ln (sin x))+ln ∣((1−cos x)/(sin x))∣}_( b) ^(π/2) J= cos b (1−ln (sin b))+ln ∣((1−cos b)/(sin b))∣; where b→0 J=1
$$\:\:\:\:{by}\:{parts} \\ $$$$\:\begin{cases}{{u}=\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)\:\Rightarrow{du}=\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:{dx}}\\{{v}=\int\:\mathrm{sin}\:{x}\:{dx}=−\mathrm{cos}\:{x}}\end{cases} \\ $$$${J}=−\mathrm{cos}\:{x}\:\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)+\int\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}\:{dx}}{\mathrm{sin}\:{x}} \\ $$$${J}=−\mathrm{cos}\:{x}\:\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)+\int\:\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}\:{dx} \\ $$$${J}=−\mathrm{cos}\:{x}\:\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)+\int\:\mathrm{cosec}\:{x}\:{dx}\:−\int\mathrm{sin}\:{xdx} \\ $$$${J}=−\mathrm{cos}\:{x}\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)+\mathrm{cos}\:{x}+\mathrm{ln}\:\mid\mathrm{cosec}\:{x}−\mathrm{cot}\:{x}\mid \\ $$$${J}=\left\{\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)\right)+\mathrm{ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\mid\right\}_{\mathrm{0}} ^{\pi/\mathrm{2}} \\ $$$${J}=\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left\{\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)\right)+\mathrm{ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\mid\right\}_{\:{b}} ^{\pi/\mathrm{2}} \\ $$$${J}=\:\mathrm{cos}\:{b}\:\left(\mathrm{1}−\mathrm{ln}\:\left(\mathrm{sin}\:{b}\right)\right)+\mathrm{ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{b}}{\mathrm{sin}\:{b}}\mid;\:{where}\:{b}\rightarrow\mathrm{0} \\ $$$${J}=\mathrm{1} \\ $$
Commented by bemath last updated on 19/Aug/20

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