∫_0 ^(π/2) ∣sin x − cos x∣dx


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Question Number 38516 by rahul 19 last updated on 26/Jun/18

$\int_{0}^{\frac{π}{2}} ∣ \sin x - \cos x ∣ d x$
Commented by math khazana by abdo last updated on 27/Jun/18
$we have cosx −sinx=(√2)cos(x+(π/4)) ⇒ I =∫_0 ^(π/2) (√2)∣cos(x+(π/4))∣dx changement x+(π/4)=t give I =(√2)∫_(π/4) ^((3π)/4) ∣cost∣ dt =(√2){ ∫_(π/4) ^(π/2) cost dt + ∫_(π/2) ^((3π)/4) −cost dt} =(√2){ [sint]_(π/4) ^(π/2) −[sint]_(π/2) ^((3π)/4) } =(√2) { 1−((√2)/2) −(((√2)/2) −1)} =(√2){ 2−(√2)) = 2(√2) −2 = I =2(√2) −2$
$w e h a v e c o s x - s i n x = \sqrt{2} c o s (x + \frac{π}{4}) \Rightarrow$ $I = \int_{0}^{\frac{π}{2}} \sqrt{2} ∣ c o s (x + \frac{π}{4}) ∣ d x c h a n g e m e n t$ $x + \frac{π}{4} = t g i v e I = \sqrt{2} \int_{\frac{π}{4}}^{\frac{3 π}{4}} ∣ c o s t ∣ d t$ $= \sqrt{2} {\int_{\frac{π}{4}}^{\frac{π}{2}} c o s t d t + \int_{\frac{π}{2}}^{\frac{3 π}{4}} - c o s t d t}$ $= \sqrt{2} {{[s i n t]}_{\frac{π}{4}}^{\frac{π}{2}} - {[s i n t]}_{\frac{π}{2}}^{\frac{3 π}{4}}}$ $= \sqrt{2} {1 - \frac{\sqrt{2}}{2} - (\frac{\sqrt{2}}{2} - 1)}$ $= \sqrt{2} {2 - \sqrt{2}) = 2 \sqrt{2} - 2 =$ $I = 2 \sqrt{2} - 2$
	Answered by MJS last updated on 26/Jun/18

	$\sin x - \cos x = - \sqrt{2} \cos (x + \frac{π}{4})$ $\cos (x + \frac{π}{4}) = 0 \Rightarrow x = \frac{π}{4}$ $\underset{0}{\int^{\frac{π}{2}}} ∣ \sin x - \cos x ∣ d x =$ $= \sqrt{2} \underset{0}{\int^{\frac{π}{4}}} \cos (x + \frac{π}{4}) d x - \sqrt{2} \underset{\frac{π}{4}}{\int^{\frac{π}{2}}} \cos (x + \frac{π}{4}) d x =$ $= 2 \sqrt{2} \underset{0}{\int^{\frac{π}{4}}} \cos (x + \frac{π}{4}) = 2 \sqrt{2} {[\sin (x + \frac{π}{4})]}_{0}^{\frac{π}{4}} =$ $= 2 \sqrt{2} - 2$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18
	$∣sinx−cosx∣ −(sinx−cosx) (Π/4)>x≥0 (sinx−cosx) (Π/2)>x≥(Π/4) see the graph from value of cosx>sinx when (Π/4)>x≥0 value of sinx>cosx when (Π/2)>x≥(Π/4) ∫_0 ^(Π/2) ∣sinx−cosx∣dx ∫_0 ^(Π/4) −(sinx−cosx)dx+∫_(Π/4) ^(Π/2) (sinx−cosx)dx ∣cosx+sinx∣_0 ^(Π/4) +∣(−cosx−sinx)∣_(Π/4) ^(Π/2) ((1/(√2))+(1/(√2))−1−0)−{(0+1)−((1/(√2))+(1/(√2)))} (2/(√2))−1−1+(2/(√2)) 2(√2) −2$
	$∣ s i n x - c o s x ∣$ $- (s i n x - c o s x) \frac{Π}{4} > x ⩾ 0$ $(s i n x - c o s x) \frac{Π}{2} > x ⩾ \frac{Π}{4}$ $s e e t h e g r a p h f r o m$ $v a l u e o f c o s x > s i n x w h e n \frac{Π}{4} > x ⩾ 0$ $v a l u e o f s i n x > c o s x w h e n \frac{Π}{2} > x ⩾ \frac{Π}{4}$ $\int_{0}^{\frac{Π}{2}} ∣ s i n x - c o s x ∣ d x$ $\int_{0}^{\frac{Π}{4}} - (s i n x - c o s x) d x + \int_{\frac{Π}{4}}^{\frac{Π}{2}} (s i n x - c o s x) d x$ $∣ c o s x + s i n x ∣_{0}^{\frac{Π}{4}} + ∣ (- c o s x - s i n x) ∣_{\frac{Π}{4}}^{\frac{Π}{2}}$ $(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 - 0) - {(0 + 1) - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}$ $\frac{2}{\sqrt{2}} - 1 - 1 + \frac{2}{\sqrt{2}}$ $2 \sqrt{2} - 2$
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