Question
$\int_{}^{} {{{\sec }^p}x\tan x\;dx = } $
therefore $\int_{}^{} {{{\sec }^p}x\tan x\,dx} = \int_{}^{} {{t^{p - 1}}dt = \frac{{{t^p}}}{p} + c} = \frac{{{{\sec }^p}x}}{p} + c.$
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$(A)$ $\int_0^1 x \cos x d x \geq \frac{3}{8}$ $(B)$ $\int_0^1 x \sin x d x \geq \frac{3}{10}$ $(C)$ $\int_0^1 x^2 \cos x d x \geq \frac{1}{2}$ $(D)$ $\int_0^1 x^2 \sin x d x \geq \frac{2}{9}$