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