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Question
Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`
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Solution
Let I = `int_0^(pi/4) (dx)/(1 + tanx)`
= `int_0^(pi/4) (dx)/(1 + sinx/cosx)`
= `int_0^(pi/4) (cos x dx)/(cosx + sinx)`
= `1/2 int_0^(pi/4) (2cosx)/(cosx + sinx) dx`
= `1/2 int_0^(pi/4) (cosx + sinx + cosx - sinx)/(cosx + sinx) dx`
= `1/2 [int_0^(pi/4) (cosx + sinx)/(cosx + sinx) dx + int_0^(pi/4) (cosx - sinx)/(cosx + sinx) dx]`
= `1/2 [int_0^(pi/4) 1dx + int_0^(pi/4) (cosx - sinx)/(cosx + sinx) dx]`
= `1/2 (I_1 + I_2)`
Where, I1 = `int_0^(pi/4) 1dx`
= `[x]_0^(pi/4) = pi/4`
And I2 = `int_0^(pi/4) (cosx - sinx)/(cosx + sinx) dx`
Let cosx + sinx = t
⇒ (–sinx + cosx)dx = dt
When x = 0, t = 1
And x = `pi/4`, t = `2/sqrt(2)`
∴ I2 = `int_1^(2/sqrt(2)) (dt)/t`
= `[logt]_1^(2/sqrt(2))`
= `log 2/sqrt(2) - log 1`
= `log 2/sqrt(2) - 0`
= `log2^(3/2)`
= `3/2 log 2`
∴ I = `1/2(I_1 + I_2)`
or I = `1/2(pi/4 + 3/2 log 2)`
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