Write a value of ^ dx - Vidyadip

Question

Write a value of $\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$

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Answer

$\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$
Let $\text{t}=\sin\text{x}\rightarrow\text{dt}=\cos\text{x dx}$
$\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}=\int\text{e}^{\log\text{t}}\text{dt}=\text{I}$
$\text{e}^{\log\text{t}}\int1\text{dt}-\Big(\int\frac{\text{de}^{\log\text{t}}}{\text{dt}}\big(\int1\text{dt}\big)\text{dt}\Big)$
$=\text{e}^{\log\text{t}}\text{t}-\Big(\int\text{e}^{\log\text{t}}\frac{1}{\text{t}}\text{t dt}\Big)$
$=\text{e}^{\log\text{t}}\text{t}-\big(\int\text{e}^{\log\text{t}}\text{dt}\big)=\text{I}$
$\rightarrow\text{e}^{\log\text{t}}\text{t}-\text{I}=\text{I}\rightarrow2\text{I}=\text{e}^{\log\text{t}}+\text{C}$
$\text{I}=\frac{1}{2}\Big[\text{te}^{\log\text{t}}\Big]+\text{C}$
Substitute back $\text{t}=\sin\text{x}$ in above expression
We get, $\text{I}=\frac{1}{2}\big[\sin{\text{x}}\text{e}^{\log\sin\text{x}}\big]+\text{C}$
$=\frac{\sin^2\text{x}}{2}+\text{C}$ $[\because\log$ with base 10 term can be changed to in (natural log) term along with a constant$]$

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