Question-78198

Question Number 78198 by john santu last updated on 15/Jan/20

Commented by john santu last updated on 15/Jan/20

d e a r M j s s i r .

i a s k f o r y o u r o p i n i o n o n t h i s a n s w e r ?

r i g h t o r n o t .

Commented by MJS last updated on 15/Jan/20

right!

Commented by john santu last updated on 15/Jan/20

t h a n k s s i r

Commented by mathmax by abdo last updated on 15/Jan/20

l e t I = \int_{0}^{\frac{π}{2}} \sqrt{1 - s i n (x)} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \sqrt{1 - c o s (\frac{π}{2} - x)} d x

=_{\frac{π}{2} - x = t} - \int_{0}^{\frac{π}{2}} \sqrt{1 - c o s t} (- d t) = \int_{0}^{\frac{π}{2}} \sqrt{1 - c o s t} d t

= \int_{0}^{\frac{π}{2}} \sqrt{2 {s i n}^{2} (\frac{t}{2})} d t = \sqrt{2} \int_{0}^{\frac{π}{2}} s i n (\frac{t}{2}) d t

= - 2 \sqrt{2} {[c o s (\frac{t}{2})]}_{0}^{\frac{π}{2}} = - 2 \sqrt{2} (\frac{\sqrt{2}}{2} - 1) = - 2 + 2 \sqrt{2} \Rightarrow

I = 2 \sqrt{2} - 2

Answered by MJS last updated on 15/Jan/20

two other possibilities

(1)

\int \sqrt{1 - \sin x} d x =

[t = \sqrt{1 - \sin x} \to d x = - \frac{2 \sqrt{1 - \sin x}}{\cos x} d x]

= \int \frac{- 2 t}{\sqrt{2 - t^{2}}} d t = 2 \sqrt{2 - t^{2}} = 2 \sqrt{1 + \sin x} + C

(2)

\int \sqrt{1 - \sin x} d x =

[t = \sin x \to d x = \frac{d t}{\cos x}]

= \int \frac{d t}{\sqrt{t + 1}} = 2 \sqrt{t + 1} = 2 \sqrt{1 + \sin x} + C

Commented by jagoll last updated on 15/Jan/20

w a w \dots a m a z i n g s i r

Commented by MJS last updated on 15/Jan/20

\dots sorry there had been a typo, now {it}^{'} s correct

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