Question

Prove That :-
$\colorbox{black}{\boxed{ \pink{ \bull\gg \gg \tt\displaystyle\tt \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| d x \ll \ll \bull}}}$

Answer with proper explanation.

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Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int\limits^{\pi/2}_{-\pi/2} | sin x| dx$

As we know that,

Property of definite integrals.

$\sf \implies \displaystyle \int\limits^{a}_{-a} f(x) dx = \int\limits^a_0 f(x) + f(-x) dx$

$\sf \implies \displaystyle 2 \int\limits^a_0 f(x)dx \ \ if \ f(-x) = f(x) \ \ ,f(x) \ is \ even$

$\sf \implies \displaystyle 0, \ if \ f(-x) = - f(x) \ \ , f(x) \ is \ odd$

Using this concept in the equation.

We can write equation as,

$\sf \implies \displaystyle 2\int\limits^{\pi/2}_{0} (sin x) dx$

As we know that,

⇒ ∫sin x dx = - cos x + c.

$\sf \implies \displaystyle 2 \bigg[ - cos \ x \bigg]_{0}^{\pi/2}$

As we know that,

In definite integration first we put upper limit then we put lower limit in the equation, we get.

$\sf \implies \displaystyle -2 \bigg[ cos \ \frac{\pi}{2} - cos(0) \bigg]$

$\sf \implies \displaystyle -2 (0 - 1). = 2.$

$\sf \implies \displaystyle \boxed{\int\limits^{\pi/2}_{-\pi/2} | sin x| dx = 2}$

                                                                                                                 

MORE INFORMATION.

Some important results.

$\sf \implies \displaystyle \int\limits^{\pi/2}_{0} ln(sin \ x) dx = \int\limits^{\pi/2}_{0} ln(cos \ x)dx = \frac{- \pi}{2} ln(2)$

$\sf \implies \displaystyle \int\limits^{\pi/2}_{0} ln(cosec \ x) dx = \int\limits^{\pi/2}_{0} ln(sec \ x)dx = \frac{ \pi}{2} ln(2)$

$\sf \implies \displaystyle \int\limits^{\pi/2}_{0} ln(tan \ x) dx = \int\limits^{\pi/2}_{0} ln(cot \ x)dx = 0$

$\sf \implies \displaystyle \int\limits^{\pi/4}_{0} ln(1 + tan \ x) dx = \frac{\pi}{8} ln(2)$

Answer 2

• please check the attached file