Help please with this integral cosx√(1+sin²x)

∫ cos(x) √(1 + sin²x) dx

u = sinx
du = cosx dx

∫ √(1 + u²) du

Use trig substitution:
u = tanθ
du = sec²θ dθ

∫ √(1 + tan²θ) sec²θ dθ
= ∫ √(sec²θ) sec²θ dθ
= ∫ secθ sec²θ dθ

Use integration by parts:
u = secθ ; dv = sec²θ dθ
du = secθtanθ dθ ; v = tanθ

Then we have:
∫ sec³θ dθ = secθtanθ - ∫ secθtan²θ dθ
∫ sec³θ dθ = secθtanθ - ∫ secθ(sec²θ - 1) dθ
∫ sec³θ dθ = secθtanθ - ∫ sec³θ - secθ dθ

Add ∫ sec³θ dθ to both sides and we have:
2 ∫ sec³θ dθ = secθtanθ + ∫ secθ dθ
2 ∫ sec³θ dθ = secθtanθ + ln|secθ + tanθ| + C
∫ sec³θ dθ = (1/2)secθtanθ + (1/2)ln|secθ + tanθ| + C

Now back substitute we said u = tanθ. This means we'll have a triangle with u on the opposite side, 1 on the adjacent side and √(u² + 1) on the hypotenuse. Plugging this in gives us:
(1/2)secθtanθ + (1/2)ln|secθ + tanθ| + C
= (1/2)√(u²+1)u + (1/2)ln|√(u² + 1) + u| + C

Plug back in x's from u's. We said u = sinx:
= (1/2)√(sin²x + 1)sinx + (1/2)ln|√(sin²x + 1) + sinx| + C <==Answer

u = sin(x)
du = cos(x)dx

cos(x)sqrt{1 + sin^2(x)}dx

= sqrt{1 + u^2} du

let u = tan v, du = sec^2 v, then 1 + u^2 = 1 + tan^2(v) = sec^2(v),

= sqrt{sec^2(v)}sec^2(v) dv

= sec^3(v) dv

you can do the rest, right?