0-pi-cos-3-x-7-sin-2-x-dx-

Question Number 149113 by bramlexs22 last updated on 03/Aug/21

\int_{0}^{π} \frac{\cos^{3} x}{7 - \sin^{2} x} dx = ?

Answered by mathmax by abdo last updated on 03/Aug/21

Ψ = \int_{0}^{π} \frac{\cos^{3} x}{7 - \sin^{2} x} dx \Rightarrow Ψ = \int_{0}^{π} \frac{\cos^{3} x}{7 - (1 - \cos^{2} x)} dx

= \int_{0}^{π} \frac{\cos^{3} x}{6 + \cos^{2} x} dx = \int_{0}^{π} \frac{cosx (\cos^{2} x + 6) - 6 cosx}{\cos^{2} x + 6} dx

= \int_{0}^{π} cosx dx - 6 \int_{0}^{π} \frac{cosx}{7 - \sin^{2} x} dx we have

\int_{0}^{π} cosx dx = {[sinx]}_{0}^{π} = 0 and

\int_{0}^{π} \frac{cosxdx}{7 - \sin^{2} x} = \int_{0}^{\frac{π}{2}} \frac{cosx}{7 - \sin^{2} x} dx + \int_{\frac{π}{2}}^{π} \frac{cosx}{7 - \sin^{2} x} (\to x = \frac{π}{2} + t)

=_{sinx = y} \int_{0}^{1} \frac{dy}{7 - y^{2}} + \int_{0}^{\frac{π}{2}} \frac{- sint dt}{7 - \cos^{2} t} (\to cost = y)

= \int_{0}^{1} \frac{dy}{7 - y^{2}} - \int_{0}^{1} \frac{dy}{7 - y^{2}} = 0 \Rightarrow Ψ = 0

Commented by bramlexs22 last updated on 03/Aug/21

yes . thanks

Commented by ArielVyny last updated on 04/Aug/21

m r m a t h m a x w h e n w e t a k e t = t a n (x)

w h a t i s t h e v a l u e o f c o s x a n d s i n x