MCQ
$\int_0^{\pi /2} {} (\sin x - \cos x)\log (\sin x + \cos x)\,dx = $
- A$ - 1$
- B$1$
- ✓$0$
- DNone of these
✓
Answer
Correct option: C.
$0$
c
(c) Put $\sin x + \cos x = t \Rightarrow - (\sin x - \cos x)dx = dt$
(c) Put $\sin x + \cos x = t \Rightarrow - (\sin x - \cos x)dx = dt$
Also as $x = 0$ to $\frac{\pi }{2},t = 1$ to $1$.
Since here limit is $'1$ to $1'$,
therefore the value of integral will be zero,
$\left\{ \because \int_{a}^{a}{f(x)dx=0} \right\}$ .
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