Question
Evaluate the following integrals:
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
✓
Answer
We know,
$\int\limits^{2\pi}_0\text{f(x)}\text{dx}=\int\limits^{2\pi}_0\text{f}(2\pi-\text{x})\text{dx}$
Hence,
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log\big(\sec(2\pi-\text{x})+\tan(2\pi-\text{x})\big)\text{dx}$
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
If
$\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec^2\text{x}-\tan^2\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(1)\text{dx}$
$2\text{I}=0$
$\text{I}=0$
$\int\limits^{2\pi}_0\text{f(x)}\text{dx}=\int\limits^{2\pi}_0\text{f}(2\pi-\text{x})\text{dx}$
Hence,
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log\big(\sec(2\pi-\text{x})+\tan(2\pi-\text{x})\big)\text{dx}$
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
If
$\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec^2\text{x}-\tan^2\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(1)\text{dx}$
$2\text{I}=0$
$\text{I}=0$
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