If $\int \frac{1}{1 + tan x} d x = A x + B ln | cos x + sin x | + C,$ then the value of ${(A + B)}^{5}$ is equal to
(where $C$ is integration constant)

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Solution

$I = \int \frac{cos x}{cos + sin x} d x = \frac{1}{2} \int \frac{2 cos x}{cos + sin x} d x = \frac{1}{2} \int \frac{(cos x + sin x) + (cos x - sin x)}{cos + sin x} d x = \frac{1}{2} \int d x + \frac{1}{2} \int \frac{cos x - sin x}{cos x + sin x} d x = \frac{x}{2} + \frac{1}{2} ln | cos x + sin x | + C \Rightarrow A = \frac{1}{2}, B = \frac{1}{2} \Rightarrow (A + B)^{5} = 1$

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