plz-help-Evalute-pi-3-pi-4-sin-2-x-1-cosx-dx-

Question Number 32928 by Cheyboy last updated on 06/Apr/18

p l z h e l p

E v a l u t e

\underset{π / 3}{\int^{π / 4}} \frac{\sin^{2} x}{\sqrt{1 - c o s x}} d x

Commented by abdo imad last updated on 06/Apr/18

l e t p u t I = \int_{\frac{π}{3}}^{\frac{π}{4}} \frac{{s i n}^{2} x}{\sqrt{1 - c o s x}} d x

I = \int_{\frac{π}{3}}^{\frac{π}{4}} \frac{4 {s i n}^{2} (\frac{x}{2}) {c o s}^{2} (\frac{x}{2})}{\sqrt{2} s i n (\frac{x}{2})} d x = \frac{4}{\sqrt{2}} \int_{\frac{π}{3}}^{\frac{π}{4}} s i n (\frac{x}{2}) {c o s}^{2} (\frac{x}{2}) d x

= \frac{4}{\sqrt{2}} \int_{\frac{π}{4}}^{\frac{π}{3}} (- s i n (\frac{x}{2})) {c o s}^{2} (\frac{x}{2}) d x

= \frac{4}{\sqrt{2}} \frac{2}{3} {[{c o s}^{3} (\frac{x}{2})]}_{\frac{π}{4}}^{\frac{π}{3}} = \frac{8}{3 \sqrt{2}} ({c o s}^{3} (\frac{π}{6}) - {c o s}^{3} (\frac{π}{8}))

= \frac{8}{3 \sqrt{2}} ({(\frac{\sqrt{3}}{2})}^{3} - {(\frac{\sqrt{2 + \sqrt{2}}}{2})}^{3})

= \frac{1}{3 \sqrt{2}} (3 \sqrt{3} - {(\sqrt{2 + \sqrt{2}})}^{3}) .

Commented by Cheyboy last updated on 07/Apr/18

t h a n k x y o u a l l f o r u r h e l p

Answered by sma3l2996 last updated on 06/Apr/18

I = \int_{π / 3}^{π / 4} \frac{{s i n}^{2} x}{\sqrt{1 - c o s x}} d x

l e t t = c o s x \Rightarrow d t = - s i n x d x

s i n x = \sqrt{1 - {c o s}^{2} x} = \sqrt{1 - t^{2}}

I = - \int_{1 / 2}^{\sqrt{2} / 2} \frac{\sqrt{1 - t^{2}}}{\sqrt{1 - t}} d t = \int_{\sqrt{2} / 2}^{1 / 2} \sqrt{1 + t} d t = \frac{2}{3} {[\sqrt{{(1 + t)}^{3}}]}_{\sqrt{2} / 2}^{1 / 2}

I = \frac{2}{3} ({(\sqrt{\frac{3}{2}})}^{3} - {(\sqrt{\frac{2 + \sqrt{2}}{2}})}^{3})

e t c \dots

Commented by Cheyboy last updated on 07/Apr/18

t h a n k y o u v r y m u c h s i r

Answered by hknkrc46 last updated on 07/Apr/18

\int_{\frac{π}{3}}^{\frac{π}{4}} \frac{\sin^{2} x \sqrt{1 - cosx}}{1 - cosx} dx = \int_{\frac{π}{3}}^{\frac{π}{4}} \frac{sinx \sqrt{1 - cosx}}{1 - cosx} sinxdx

= \int_{\frac{π}{3}}^{\frac{π}{4}} \frac{\sqrt{1 - \cos^{2} x} \sqrt{1 - cosx}}{1 - cosx} sinxdx

★ 1 - cosx = u \Rightarrow sinxdx = du \land (x \to \frac{π}{4} \Rightarrow u \to \frac{2 - \sqrt{2}}{2} \land x \to \frac{π}{3} \Rightarrow u \to \frac{1}{2})

= \int_{\frac{1}{2}}^{\frac{2 - \sqrt{2}}{2}} \frac{\sqrt{1 - {(1 - u)}^{2}} \sqrt{u}}{u} du = \int_{\frac{1}{2}}^{\frac{2 - \sqrt{2}}{2}} \frac{\sqrt{2 u - u^{2}}}{\sqrt{u}} du

= \int_{\frac{1}{2}}^{\frac{2 - \sqrt{2}}{2}} \sqrt{2 - u} du = - \frac{2}{3} {(2 - u)}^{\frac{3}{2}} ∣_{\frac{1}{2}}^{\frac{2 - \sqrt{2}}{2}}

= - \frac{2}{3} [{(2 - \frac{2 - \sqrt{2}}{2})}^{\frac{3}{2}} - {(2 - \frac{1}{2})}^{\frac{3}{2}}]

= - \frac{2}{3} [{(\frac{2 + \sqrt{2}}{2})}^{\frac{3}{2}} - {(\frac{3}{2})}^{\frac{3}{2}}]

Commented by Cheyboy last updated on 07/Apr/18

t h a n k u v e r y m u c h s i r

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