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
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