$\int_{0}^{\frac{π}{8}} \frac{s e c^{2} 2 x}{2} d x =$

\frac{1}{4}

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\frac{1}{3}

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\frac{1}{2}

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N o n e o f t h e s e

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Solution

The correct option is A $\frac{1}{4}$
$\frac{1}{2} \int_{0}^{\frac{π}{8}} s e c^{2} 2 x d x = \frac{1}{4} [t a n 2 x]_{0}^{\frac{π}{8}} = \frac{1}{4} [1] = \frac{1}{4} .$

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Similar questions

\int_{0}^{\frac{π}{8}} \frac{s e c^{2} 2 x}{2} d x =

\int_{0}^{π / 2} \frac{d x}{1 + {tan}^{5} x} = \frac{π}{4}

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a + x) d x = \int_{0}^{π / 2} \frac{d x}{1 + {tan}^{3} x} = \int_{0}^{π / 2} \frac{d_{X}}{1 + {cot}^{3} x} = \frac{π}{4}

\int_{0}^{π / 2} \frac{d_{X}}{4 {cos}^{2} x + 9 {sin}^{2} x} =

\int_{0}^{\frac{π}{2}} \frac{d x}{2 + c o s x} =

\int_{0}^{\frac{π}{2}} \frac{d x}{1 + tan x} =

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