∫_0^π/4tan^2 x dx=∫_0^π/4(sec^2 x 1)dx =∫_0^π/4sec^2 xdx ∫_0^π/41dx=[tan x]_0^π/4 [x]_0^π/4=1 π/4 ...

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First Fundamental Theorem of Calculus

∫0π/4tan2 x d...

Question

$\int_{0}^{\frac{π}{4}} t a n^{2} x d x =$

1 - \frac{π}{4}

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1 + \frac{π}{4}

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\frac{π}{4} - 1

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\frac{π}{4}

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Solution

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

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$\int_{0}^{\frac{π}{2}} (c o s x - s i n x) e^{x} d x =$

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