find-pi-2-pi-2-cosx-cos-3-x-dx-

Question Number 53465 by maxmathsup by imad last updated on 22/Jan/19

f i n d \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{c o s x - {c o s}^{3} x} d x

Commented by maxmathsup by imad last updated on 23/Jan/19

l e t I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{c o s x - {c o s}^{3}} d x \Rightarrow I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{c o s x} \sqrt{1 - {c o s}^{2} x} d x

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{c o s x} ∣ s i n x ∣ d x = 2 \int_{0}^{\frac{π}{2}} \sqrt{c o s x} s i n x d x

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

= - 2 {[\frac{1}{1 + \frac{1}{2}} {c o s}^{1 + \frac{1}{2}} x]}_{0}^{\frac{π}{2}} = - 2 . \frac{2}{3} {[{c o s}^{\frac{3}{2}} x]}_{0}^{\frac{π}{2}} = - \frac{4}{3} (- 1) = \frac{4}{3}

Answered by ajfour last updated on 22/Jan/19

\int_{- π / 2}^{π / 2} \sqrt{\cos x} ∣ \sin x ∣ d x

= - 2 \int_{0}^{π / 2} \sqrt{\cos x} (\sin x) d x

= - 2 \int_{1}^{0} \sqrt{t} d t = \frac{4}{3} .

Commented by malwaan last updated on 23/Jan/19

t = cosx

\Rightarrow dt = - sinxdx

where is (-) ?

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