Question
Evaluate the following integrals:
$\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
$\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
✓
Answer
Let $\text{I}=\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
Let $\log\text{x}=\text{z}$
$\Rightarrow\text{x = e}^{\text{z}}$
$\Rightarrow\text{dx}=\text{e}^{\text{z}}\text{dz}$
$\therefore\text{I}=\int\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{e}^{\text{z}}\text{dz}$
Here, $\text{f(z)}=\tan\text{z}$ and $\text{f}'\text{(z)}=\sec^2\text{z}$
And we know that
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x) + C}$
$\therefore\int\text{e}^{\text{z}}\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{dz}=\text{e}^{\text{z}}\tan\text{z + C}$
$\therefore\text{I}=\text{x}\tan(\log\text{x})+\text{C}$
Let $\log\text{x}=\text{z}$
$\Rightarrow\text{x = e}^{\text{z}}$
$\Rightarrow\text{dx}=\text{e}^{\text{z}}\text{dz}$
$\therefore\text{I}=\int\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{e}^{\text{z}}\text{dz}$
Here, $\text{f(z)}=\tan\text{z}$ and $\text{f}'\text{(z)}=\sec^2\text{z}$
And we know that
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x) + C}$
$\therefore\int\text{e}^{\text{z}}\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{dz}=\text{e}^{\text{z}}\tan\text{z + C}$
$\therefore\text{I}=\text{x}\tan(\log\text{x})+\text{C}$
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