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

 ∫_0 ^( π)  ((cos ^3 x)/(7−sin ^2 x)) dx =?

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

Ψ=∫_0 ^(π )  ((cos^3 x)/(7−sin^2 x)) dx ⇒Ψ=∫_0 ^π  ((cos^3 x)/(7−(1−cos^2 x)))dx  =∫_0 ^π  ((cos^3 x)/(6+cos^2 x))dx =∫_0 ^π  ((cosx(cos^2 x+6)−6cosx)/(cos^2 x +6))dx  =∫_0 ^π  cosx dx−6∫_0 ^π  ((cosx)/(7−sin^2 x))dx  we have  ∫_0 ^π  cosx dx=[sinx]_0 ^π =0  and  ∫_0 ^π  ((cosxdx)/(7−sin^2 x))=∫_0 ^(π/2)  ((cosx)/(7−sin^2 x))dx +∫_(π/2) ^π  ((cosx)/(7−sin^2 x))(→x=(π/2)+t)  =_(sinx=y)    ∫_0 ^1  (dy/(7−y^2 )) +∫_0 ^(π/2)  ((−sint dt)/(7−cos^2 t))(→cost=y)  =∫_0 ^1  (dy/(7−y^2 ))  −∫_0 ^1  (dy/(7−y^2 )) =0 ⇒Ψ=0

Commented bybramlexs22 last updated on 03/Aug/21

yes. thanks

Commented byArielVyny last updated on 04/Aug/21

mr mathmax when we take t=tan(x)  what is the value of cosx and sinx