-0-4- x-4 x2 ln 1-tan x d x-

The correct option is C π^3/192 In 2Let I = ∫_0^π/44x (π/4 x)In (1+tan x)dx = ∫_0^π/44 (π/4 x)x In(1 + tan(π/4 x))dx = 4 In 2 ∫_0^π/4x (π/4 x)dx I ⇒ I ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

∫0π/4π x-4 x2...

Question

$\int_{0}^{\frac{π}{4}} (π x - 4 x^{2})$ In(1 + tan x)dx = ___

\frac{π^{3}}{192}

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\frac{π^{3}}{192} I n 2

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\frac{π^{2}}{96}

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\frac{π^{2}}{96} I n 2

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Solution

The correct option is B $\frac{π^{3}}{192} I n 2$
Let $I = \int_{0}^{\frac{π}{4}} 4 x (\frac{π}{4} - x) I n (1 + t a n x) d x = \int_{0}^{\frac{π}{4}} 4 (\frac{π}{4} - x) x I n (1 + t a n (\frac{π}{4} - x)) d x = 4 I n 2 \int_{0}^{\frac{π}{4}} x (\frac{π}{4} - x) d x - I \Rightarrow I = 2 I n 2 \int_{0}^{\frac{π}{4}} x (\frac{π}{4} - x) d x = \frac{π^{3}}{192} I n 2$

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∫π40(πx−4x2) In(1 + tan x)dx = ___

The correct option is B π3192In2Let I=∫π404x(π4−x)In(1+tanx)dx=∫π404(π4−x)x In(1+tan(π4−x))dx=4 In 2∫π40x(π4−x)dx−I⇒I=2In2∫π40x(π4−x)dx=π3192In2

$\int_{0}^{\frac{π}{4}} (π x - 4 x^{2})$ In(1 + tan x)dx = ___