Mathematics

$\int \sin x \log (\cos x)dx$

SOLUTION
Let $t=\cos{x}\Rightarrow dt=-\sin{x}dx$
$\displaystyle\int{\sin{x}\log{\cos{x}}dx}$
$=-\displaystyle\int{\log{t}dt}$
Integrating by parts,
Let $u=\log{t}\Rightarrow du=\dfrac{1}{t}dt$
$dv=dt\Rightarrow v=t$
$=-\left[t\log{t}-\displaystyle\int{t\times\dfrac{dt}{t}}\right]$
$=-t\log{t}+\displaystyle\int{dt}$
$=-t\log{t}+t+c$ where $c$ is the constant of integration
$=-\cos{x}\log{\left(\cos{x}\right)}+\cos{x}+c$ where $t=\cos{x}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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