Evaluation of definite integrals

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Question

$\int_{0}^{π / 4} (π x - 4 x^{2}) \log (1 + \tan x) d x =$

Moderate

Solution

$\begin{array}{l} Let I = \int_{0}^{π / 4} 4 x (\frac{π}{4} - x) \log (1 + \tan x) d x \to (1). \\ Apply \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \\ \Rightarrow I = \int_{0}^{π / 4} 4 (\frac{π}{4} - x) x \log (1 + \tan (\frac{π}{4} - x)) d x \to (2) \\ (1) + (2) \Rightarrow 2 I = (\log 2) \cdot 4 \int_{0}^{π / 4} x (\frac{π}{4} - x) d x \\ \Rightarrow I = 2 \log 2 {(\frac{π}{4} \cdot \frac{x^{2}}{2} - \frac{x^{3}}{3})}_{0}^{π / 4} \\ = 2 \log 2 (\frac{π}{8} \frac{π^{2}}{16} - \frac{1}{3} \frac{π^{3}}{64}) \end{array}$
$= \frac{π^{3}}{192} . \log 2$

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Evaluation of definite integrals

Remember concepts with our Masterclasses.

∫0π/4πx−4x2log1+tanxdx=

Let I=∫0π/4 4xπ4−xlog⁡(1+tan⁡x)dx→ (1). Apply ∫ab f(x)dx=∫ab f(a+b−x)dx ⇒I=∫0π/4 4π4−xxlog⁡1+tan⁡π4−xdx→(2)(1)+(2)⇒2I=(log⁡2)⋅4∫0π/4 xπ4−xdx⇒I=2log⁡2π4⋅x22−x330π/4 =2log⁡2π8π216−13π364 =π3192.log2

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$\int_{0}^{π / 4} (π x - 4 x^{2}) \log (1 + \tan x) d x =$