int_{0}^{pi / 2} (sin x - cos x) log(sin x + | vedclass

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Question

(C) Let $I = \int_{0}^{\pi / 2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx$. Consider the substitution $t = \sin x + \cos x$. Then $dt = (\cos x -

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $I = \int_{0}^{\pi / 2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx$. Consider the substitution $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) \, dx$,which implies $-( \sin x - \cos x) \, dx = dt$. Now,change the limits of integration: When $x = 0$,$t = \sin(0) + \cos(0) = 0 + 1 = 1$. When $x = \frac{\pi}{2}$,$t = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1$. Substituting these into the integral,we get: $I = \int_{1}^{1} - \log(t) \, dt$. Since the lower limit and upper limit are the same,the value of the definite integral is $0$,because $\int_{a}^{a} f(t) \, dt = 0$.

$\int_{0}^{\pi / 2} (\sin x - \cos x) \log(\sin x + \cos x) \, dx = $

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