pi-2-pi-2-cos-x-cos-3-x-dx-

Question Number 91479 by Zainal Arifin last updated on 01/May/20

\underset{- π / 2}{\int^{π / 2}} \sqrt{\cos x - \cos^{3} x} dx = \dots

Commented by jagoll last updated on 01/May/20

\sqrt{\cos x (1 - \cos^{2} x)} = \sqrt{\cos x} \sqrt{\sin^{2} x}

\underset{- \frac{π}{2}}{\int^{0}} \sqrt{\cos x} (- \sin x) d x + \underset{0}{\int^{\frac{π}{2}}} \sqrt{\cos x} \sin x d x

n o w e a s y t o s o l v e

Commented by Zainal Arifin last updated on 01/May/20

Ok . thanks

Commented by Prithwish Sen 1 last updated on 01/May/20

\int \sqrt{cosx} sinx dx put cosx = t^{2} \Rightarrow - sinxdx = 2 tdt

= - \int t . 2 t . dt = - \frac{2}{3} . {(cosx)}^{\frac{3}{2}}

\int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{cosx} . sinxdx = 2 \int_{0}^{\frac{π}{2}} \sqrt{cosx} . sinxdx = 2 {[- \frac{2}{3} {(cosx)}^{\frac{2}{3}}]}_{0}^{\frac{π}{2}}

= 2 . [- (0 - \frac{2}{3})] = \frac{4}{3} please check .

Commented by jagoll last updated on 01/May/20

\sqrt{\sin^{2} x} \neq \sin x

\sqrt{\sin^{2} x} = ∣ \sin x ∣

Answered by john santu last updated on 01/May/20

Commented by Prithwish Sen 1 last updated on 01/May/20

I think in 0 ⩽ x ⩽ \frac{π}{2} sinx and cosx ⩾ 0

sorry fix it .

Commented by jagoll last updated on 01/May/20

0 ⩽ x ⩾ \frac{π}{2} = 0 ⩽ x ⩽ \frac{π}{2} ?

Commented by Prithwish Sen 1 last updated on 01/May/20

Thank you sir .