0-2-sinx-cosx-please-help-me-out-

Question Number 152088 by rexford last updated on 25/Aug/21

\int_{0}^{\frac{Π}{2}} ∣ s i n x - c o s x ∣

p l e a s e h e l p m e o u t

Answered by Olaf_Thorendsen last updated on 25/Aug/21

I = \int_{0}^{\frac{π}{2}} ∣ \sin x - \cos x ∣ d x

I = \int_{0}^{\frac{π}{2}} ∣ \sqrt{2} \sin (\frac{π}{4} - x) ∣ d x

I = 2 \int_{0}^{\frac{π}{4}} \sqrt{2} \sin (\frac{π}{4} - x) d x

I = 2 \sqrt{2} {[\cos (\frac{π}{4} - x)]}_{0}^{\frac{π}{4}}

I = 2 \sqrt{2} (1 - \frac{1}{\sqrt{2}}) = 2 (\sqrt{2} - 1)

Answered by puissant last updated on 25/Aug/21

= \int_{0}^{\frac{π}{4}} (c o s x - s i n x) d x + \int_{\frac{π}{4}}^{\frac{π}{2}} (s i n x - c o s x) d x

= {[s i n x + c o s x]}_{0}^{\frac{π}{4}} + {[- c o s x - s i n x]}_{\frac{π}{4}}^{\frac{π}{2}}

= \sqrt{2} - 1 + (- 1 + \sqrt{2})

= 2 \sqrt{2} - 2

= 2 (\sqrt{2} - 1)

Commented by peter frank last updated on 25/Aug/21

good

Commented by rexford last updated on 26/Aug/21

t h a n k y o u

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