The correct option is A 1 π4 tan^2x = sec^2x 1 ⇒ ∫_0^π/4 1 #160; dx #160; + #160; #160; #160; ∫_0^π/4 (sec^2x) #160; d ...

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Definite Integral as Limit of Sum

∫ 0 π /4 tan ...

Question

$\int_{0}^{π / 4} {tan}^{2} x d x =$

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} - 1

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Solution

The correct option is A $1 - \frac{π}{4}$
$t a n^{2} x = s e c^{2} x - 1$

$\Rightarrow$ $\int_{0}^{π / 4} - 1 d x$ + $\int_{0}^{π / 4} (s e c^{2} x) d x$

= $- π / 4$ + [ $t a n x$ } $_{0}^{π / 4}$

= 1 - $π / 4$

So, the correct option is 'A'.

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Similar questions

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

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

\int_{0}^{π / 2} \frac{1}{1 + {tan}^{3} x} d x = \frac{π}{4}

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

$\int_{0}^{\frac{π}{4}} t a n^{2} x d x =$ [Roorkee 1983, Pb. CET 2000]

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