Evaluate:\(\int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan x }\)

1.	Evaluate:\(\int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan x }\)
Answer» Let \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan x }\) ⇒ \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan (\frac {\pi}4- x) }\) \(\left(\because \int\limits_a^b f(x) dx = \int\limits^b_af(a + b - x)dx\right)\) ⇒ \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + \cfrac{tan\frac{\pi}4 - tanx}{1 + tan\frac{\pi}{4}tanx} }\) \(= \int\limits_{0}^{\frac{\pi}4} \frac{dx}{ \cfrac{2 + tanx + 1 - tanx}{1 + tanx}}\) \(\left(\because tan \frac{\pi}{4} = 1\right)\) \(= \int\limits_{0}^{\frac{\pi}4} \frac{1 + tan x }{2}dx\) \(= \frac12 [x + log\, secx] ^{\frac{\pi}4}_0\) \((\because \int tan x \,dx = log\,secx)\) \(= \frac 12 \left[\left(\frac{\pi}{4} - 0\right) + log \left(sec\frac{\pi}{4}\right) - log\,sec 0\right]\) \(= \frac12 \left(\frac{\pi}{4} + log \sqrt 2 - log \,1\right)\) \(= \frac 12 \left( \frac{\pi}{4} + \frac12 log2\right)\) \((\because log 1 = 0)\) \(= \frac{\pi}6 + \frac14 log2.\)

Answer»

Let \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan x }\)

⇒ \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + tan (\frac {\pi}4- x) }\)

\(\left(\because \int\limits_a^b f(x) dx = \int\limits^b_af(a + b - x)dx\right)\)

⇒ \(I = \int\limits_{0}^{\frac{\pi}4} \frac{dx}{1 + \cfrac{tan\frac{\pi}4 - tanx}{1 + tan\frac{\pi}{4}tanx} }\)

\(= \int\limits_{0}^{\frac{\pi}4} \frac{dx}{ \cfrac{2 + tanx + 1 - tanx}{1 + tanx}}\) \(\left(\because tan \frac{\pi}{4} = 1\right)\)

\(= \int\limits_{0}^{\frac{\pi}4} \frac{1 + tan x }{2}dx\)

\(= \frac12 [x + log\, secx] ^{\frac{\pi}4}_0\) \((\because \int tan x \,dx = log\,secx)\)

\(= \frac 12 \left[\left(\frac{\pi}{4} - 0\right) + log \left(sec\frac{\pi}{4}\right) - log\,sec 0\right]\)

\(= \frac12 \left(\frac{\pi}{4} + log \sqrt 2 - log \,1\right)\)

\(= \frac 12 \left( \frac{\pi}{4} + \frac12 log2\right)\) \((\because log 1 = 0)\)

\(= \frac{\pi}6 + \frac14 log2.\)

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