Question

Tolong dibantu iya kak

Answer 1

Verified answer

₀∫⁹⁰ (cos x)/(1 + sin² x) dx

misal

u = sin x

du/dx = cos x

dx = du/cos x

∫ (cos x)/(1 + u²) . du/(cos x)

= ∫ 1/(1 + u²) du

Misal

u = tan θ

du/dθ = sec² θ

du = sec² θ dθ

tan θ = u

θ = arc tan u

∫ 1/(1 + u²) du

= ∫ 1/(1 + tan² θ) . sec² θ dθ

= ∫ 1/(sec² θ) . sec² θ dθ

= ∫ 1 dθ

= θ + C

= arc tan u + C

= arc tan (sin x) + C

Jadi

₀∫⁹⁰ (cos x)/(1 + sin² x) dx

= arc tan (sin x) ₀|⁹⁰

= arc tan (sin π/2) - arc tan (sin 0)

= arc tan (1) - arc tan 0

= π/4 - 0

= π/4