Question

π/2prove that: ∫ log cosx dx= - π/2 log 2 0

Answer 1

Step-by-step explanation:

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Answer 2

Given:

$\int_{0}^{\frac{\pi}{2}} \log \cos (x) d x=-\frac{\pi}{2} \log 2$

Step-by-step explanation:

Let us consider the L.H.S as ‘I’.

$I=\int_{0}^{\frac{\pi}{2}} \log \cos (x) d x$

In this problem, we need to follow the below identities for simplified calculation.

$\int_{0}^{\frac{\pi}{2}} \log \cos x d x=\int_{0}^{\frac{\pi}{2}} \log \sin x d x$

$\int_{0}^{\pi} \log \sin x d x=2 \int_{0}^{\frac{\pi}{2}} \log \sin x d x$

Now, the integral becomes,

$2 I=\int_{0}^{\frac{\pi}{2}} \log \cos x d x+\int_{0}^{\frac{\pi}{2}} \log \sin x d x$

$I=\int_{0}^{\frac{\pi}{2}} (\log \cos x+\log \sin x) d x$

$\bold{[\because \log a b=\log a+\log b]}$

$I=\int_{0}^{\pi / 2} \log (\sin x \cdot \cos x) d x$

$\bold{[\because \cos (x) \sin (x)=\frac{\sin (2 x)}{2}]}$

$I=\int_{0}^{\frac{\pi}{2}} \log \frac{\sin 2 x}{2} d x$

$\bold{\left[\because \log \frac{a}{b}=\log a-\log b\right]}$

$I=\int_{0}^{\frac{\pi}{2}} \log \sin 2 x-\log 2 d x$

$I=\int_{0}^{\frac{\pi}{2}} \log \sin 2 x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x$

$I=\int_{0}^{\frac{\pi}{2}} (\log \sin 2 x) d x-\frac{\pi \log 2}{2}$

Let $u = 2x;$

$\frac{du}{dx}=2 \Rightarrow d x=\frac{d u}{2}$

If $x=0 \Rightarrow u=0$

If $x=\frac{\pi}{2} \Rightarrow u=\pi$

Now,

$2 I=\int_{0}^{\pi} \frac{1}{2} \log \sin u d u-\frac{\pi \log 2}{2}$

$2 I=\frac{1}{2} \int_{0}^{\pi} \log \sin u d u-\frac{\pi \log 2}{2}$

$\bold{[\because 2 I=2 \int_{0}^{\frac{\pi}{2}} \ln \sin x d x=\int_{0}^{\pi} \ln \sin x d x=\int_{0}^{\pi} \ln \sin u d u]}$

$2 I=\frac{1}{2} 2 I-\frac{\pi \log 2}{2}$

$2 I=I-\frac{\pi \log 2}{2}$

$I=\left[-\frac{\pi \log 2}{2}\right]$

Hence proved.