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# Integrate : $\int \log (\sin x ) \cos x dx$

$\begin {array} {1 1} (a)\;\cos x [\log (\sin x) -1]+c \\ (b)\;\sin x [\log (\sin x) -1]+c \\ (c)\;\cos x [\log (\sin x) +1]+c \\ (d)\;None \end {array}$
Can you answer this question?

$\sin x=t$
=> $\cos x dx=dt$
$\int \log t .dt$
=> $[\log t.t-t]+c$
=> $t [\log t+1]+c$
=> $\sin x [ \log \sin x -1 ]+c$
Hence b is the correct answer.
answered Dec 19, 2013 by