Question
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
$\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
✓
Answer
Note: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
Putting $\log(\sec\text{x}+\tan\text{x})=\text{t}$
$\Rightarrow\frac{\sec\text{x}\tan\text{x}+\sec^2\text{x}}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x}\frac{(\sec\text{x}+\tan\text{x})}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x }\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\sec\text{x}+\tan\text{x})|+\text{C}$
Let $\text{I}=\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
Putting $\log(\sec\text{x}+\tan\text{x})=\text{t}$
$\Rightarrow\frac{\sec\text{x}\tan\text{x}+\sec^2\text{x}}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x}\frac{(\sec\text{x}+\tan\text{x})}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x }\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\sec\text{x}+\tan\text{x})|+\text{C}$
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