Question
Evaluate the following integrals:
$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
✓
Answer
$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
$\text{Let }\text{ log}(\sec\text{x}+ \tan\text{x})=\text{t}$
$\Rightarrow \frac{(\sec\text{x} \tan\text{x}+\sec^{2}\text{x})} {(\sec\text{x} +\tan\text{x})}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\sec\text{x} (\sec\text{x}+\tan\text{x})} {(\sec\text{x} +\tan\text{x})}\text{dx}=\text{dt}$
$\text{Now},\int\sec \text{x}.\text{log}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int \text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{[\text{log}(\sec\text{x}+\tan\text{x})]^2}{2}+\text{C}$
$\text{Let }\text{ log}(\sec\text{x}+ \tan\text{x})=\text{t}$
$\Rightarrow \frac{(\sec\text{x} \tan\text{x}+\sec^{2}\text{x})} {(\sec\text{x} +\tan\text{x})}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\sec\text{x} (\sec\text{x}+\tan\text{x})} {(\sec\text{x} +\tan\text{x})}\text{dx}=\text{dt}$
$\text{Now},\int\sec \text{x}.\text{log}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int \text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{[\text{log}(\sec\text{x}+\tan\text{x})]^2}{2}+\text{C}$
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