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Integrate log(sinx+cosx)dx from -pi/4 to pi/4.

First put x = -x and add \(\begin{array}{l}I = \int_{-\pi/4}^{\pi/4}log(sinx+cosx)dx = \int_{-\pi/4}^{\pi/4}log(-sinx+cosx)dx 2\\I = \int_{-\pi/4}^{\pi/4}log(-sin^2x+cos^2x)dx = \int_{-\pi/4}^{\pi/4}log(cos2x)dx = 2\int_{0}^{\pi/4}log(cos2x)dx\\I =\int_{0}^{\pi/4}log(cos2x)dx... View Article

Integrate sqrt(sinx) with limit 0 to pi/2.

∫√sin(x)dx Rewrite/simplify using trigonometric/hyperbolic identities: =∫√2cos^2(2x−π) / [4])−1dx =∫√1−2sin^2(2x−π) / [4])dx Substitute u=(2x−π) / [4])⟶ du/dx=1/ 2 ⟶ dx=2du: =2∫√1−2sin^2(u)du... View Article

Integrate log(1+tan x) from 0 to pi/4.

Answer: Let us consider \(\begin{array}{l}I = \int_0^{\pi/4} log(1+tanx)dx\end{array} \) We know that ⇒ \(\begin{array}{l}\int_0^af(x)dx=\int_0^af(a-x)dx\end{array} \) ⇒ \(\begin{array}{l}I = \int_0^{\pi/4}log(1+tan[{\frac{\pi}{4}}-x]dx\end{array} \)... View Article