∫_(−1/2) ^(1/2) ∣ x cos(((πx)/2))∣ dx =


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Question Number 82474 by zainal tanjung last updated on 21/Feb/20

$\underset{- 1 / 2}{\int^{1 / 2}} ∣ x \cos (\frac{π x}{2}) ∣ d x =$
Commented by mathmax by abdo last updated on 21/Feb/20
$let I =∫_(−(1/2)) ^(1/2) ∣xcos(((πx)/2))∣dx ⇒I =2∫_0 ^(1/2) ∣x∣∣cos(((πx)/2))∣dx changement ((πx)/2) =t give I=2 ∫_0 ^(π/4) (2/π)t cost ×(2/π)dt =(8/π^2 ) ∫_0 ^(π/4) tcost dt =_(by parts) (8/π^2 ){ [tsint]_0 ^(π/4) −∫_0 ^(π/4) sint dt} =(8/π^2 ){(π/(4(√2))) +[cost]_0 ^(π/4) } =(8/π^2 ){(π/(4(√2))) +(1/(√2))−1}$
$l e t I = \int_{- \frac{1}{2}}^{\frac{1}{2}} ∣ x c o s (\frac{π x}{2}) ∣ d x \Rightarrow I = 2 \int_{0}^{\frac{1}{2}} ∣ x ∣∣ c o s (\frac{π x}{2}) ∣ d x$ $c h a n g e m e n t \frac{π x}{2} = t g i v e I = 2 \int_{0}^{\frac{π}{4}} \frac{2}{π} t c o s t \times \frac{2}{π} d t$ $= \frac{8}{π^{2}} \int_{0}^{\frac{π}{4}} t c o s t d t =_{b y p a r t s} \frac{8}{π^{2}} {{[t s i n t]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} s i n t d t}$ $= \frac{8}{π^{2}} {\frac{π}{4 \sqrt{2}} + {[c o s t]}_{0}^{\frac{π}{4}}} = \frac{8}{π^{2}} {\frac{π}{4 \sqrt{2}} + \frac{1}{\sqrt{2}} - 1}$
Commented by zainal tanjung last updated on 23/Apr/20

$thank sir$
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