Question
Evaluate the following integrals:
$\int\sin\text{x}\log(\cos\text{x})\text{dx}$
✓
Answer
Let $\text{I}=\int\sin\text{x}\cdot\log(\cos\text{x})\text{dx}$
Let $\cos\text{x = t}$
$\Rightarrow-\sin\text{x dx =}\text{ dt}$
$\Rightarrow\sin\text{x dx =}-\text{dt}$
$\therefore\text{I}=-\int\log\text{t dt}$
$=-\int1\cdot\log\text{t dt}$
Taking log t as the first function and 1 as the second function.
$=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t})\int1\text{dt}\big\}\text{dt}$
$=-[\log\text{t}\cdot\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}]$
$=-[\log\text{t}\cdot\text{t}-\text{t}]+\text{C}$
$=-\text{t}(\log\text{t}-1)+\text{C} \dots(1)$
Substituting the value of t in eq (1)
$=-\cos\text{x}\{\log(\cos\text{x})-1\}+\text{C}$
$=\cos\text{x}\{1-\log(\cos\text{x})\}+\text{C}$
Let $\cos\text{x = t}$
$\Rightarrow-\sin\text{x dx =}\text{ dt}$
$\Rightarrow\sin\text{x dx =}-\text{dt}$
$\therefore\text{I}=-\int\log\text{t dt}$
$=-\int1\cdot\log\text{t dt}$
Taking log t as the first function and 1 as the second function.
$=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t})\int1\text{dt}\big\}\text{dt}$
$=-[\log\text{t}\cdot\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}]$
$=-[\log\text{t}\cdot\text{t}-\text{t}]+\text{C}$
$=-\text{t}(\log\text{t}-1)+\text{C} \dots(1)$
Substituting the value of t in eq (1)
$=-\cos\text{x}\{\log(\cos\text{x})-1\}+\text{C}$
$=\cos\text{x}\{1-\log(\cos\text{x})\}+\text{C}$
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