Blog: `int cos(x) ln(sin(x)) dx` Evaluate the integral

Sunday, May 17, 2009

$\displaystyle\int{\cos{{\left({x}\right)}}}{\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

You need to use the substitution $\displaystyle{\sin{{x}}}={t},$ such that:

$\displaystyle{\sin{{x}}}={t}\Rightarrow{\cos{{x}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}$

Replacing the variable, yields:

$\displaystyle\int{\cos{{x}}}\cdot{\ln{{\left({\sin{{x}}}\right)}}}{\left.{d}{x}\right.}=\int{\ln{{t}}}{\left.{d}{t}\right.}$

You need to use the integration by parts such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={\ln{{t}}}\Rightarrow{d}{u}=\frac{{{\left.{d}{t}\right.}}}{{t}}$

$\displaystyle{d}{v}={1}\Rightarrow{v}={t}$

$\displaystyle\int{\ln{{t}}}{\left.{d}{t}\right.}={t}\cdot{\ln{{t}}}-\int{t}\cdot\frac{{{\left.{d}{t}\right.}}}{{t}}$

$\displaystyle\int{\ln{{t}}}{\left.{d}{t}\right.}={t}\cdot{\ln{{t}}}-\int{\left.{d}{t}\right.}$

$\displaystyle\int{\ln{{t}}}{\left.{d}{t}\right.}={t}\cdot{\ln{{t}}}-{t}+{C}$

Replacing back the variable, yields:

$\displaystyle\int{\cos{{x}}}\cdot{\ln{{\left({\sin{{x}}}\right)}}}{\left.{d}{x}\right.}={\sin{{x}}}\cdot{\ln{{\left({\sin{{x}}}\right)}}}-{\sin{{x}}}+{C}$

Hence, evaluating the integral, using substitution, then integration by parts, yields $\displaystyle\int{\cos{{x}}}\cdot{\ln{{\left({\sin{{x}}}\right)}}}{\left.{d}{x}\right.}={\sin{{x}}}\cdot{\left({\ln{{\left({\sin{{x}}}\right)}}}-{1}\right)}+{C}.$

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Sunday, May 17, 2009

$\displaystyle\int{\cos{{\left({x}\right)}}}{\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

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