Evaluate ∫13+sin2xdx
122tan−13tan x+122+C
122tan−1tan x+122+C
162tan−13tan x+122+C
12tan−13tan x+122+C
I=∫13+sin2xdx=∫13sin2x+cos2x+2sinxcosxdx=∫sec2x3tan2x+2tanx+3dx[Dividing Nr and Dr by cos2 x] Putting tan x = t and sec2 x dx = dt, we get I=∫dt3t2+2t+3=13∫dtt2+23t+1=13∫dtt+132+2232I=131223tan−1t+13223+C=122tan−13t+122+C=122tan−13tan x+122+C