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$\int_0^{\pi / 6} \frac{\sin x}{\cos ^3 x} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{6}$
Let $I=\int_0^{\pi / 6} \frac{\sin x}{\cos ^3 x} d x=\int_0^{\pi / 6} \tan x \sec ^2 x d x$
Put $t=\tan x \Rightarrow d t=\sec ^2 x d x$, then we have
$I=\int_0^{\frac{1}{\sqrt{3}}} t d t=\left[\frac{t^2}{2}\right]_0^{\sqrt{3}}=\frac{1}{6}$
Put $t=\tan x \Rightarrow d t=\sec ^2 x d x$, then we have
$I=\int_0^{\frac{1}{\sqrt{3}}} t d t=\left[\frac{t^2}{2}\right]_0^{\sqrt{3}}=\frac{1}{6}$
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