Integrate (sin^2(t)+1)/(cos^4(t))dt
To integrate the given expression, we can use a substitution method. Let's use the substitution u = cos(t).
To find du, we need to differentiate u with respect to t. Since u = cos(t), we can differentiate both sides of the equation with respect to t:
du/dt = d/dt (cos(t))
du/dt = -sin(t)
Now, let's solve for dt in terms of du:
dt = du / (-sin(t))
Substituting u = cos(t) and dt = du / (-sin(t)) into the integral expression, we get:
∫ [(sin^2(t) + 1) / (cos^4(t))] dt
= ∫ [(sin^2(t) + 1) / (cos^4(t))] (-sin(t)) du
Now, we simplify the integrand:
= ∫ [(-sin(t) / cos^4(t)) + (1 / cos^4(t))] sin(t) du
= ∫ [(-sin(t) / cos^4(t))sin(t) + (1 / cos^4(t))sin(t)] du
= ∫ [-sin^2(t) / cos^4(t) + sin(t) / cos^4(t)] du
Using the substitution u = cos(t):
= ∫ [-1/u^4 + du/u^3]
= ∫ [-1/u^4] du + ∫ [du/u^3]
= (1/3u^3) - (1/u) + C
Finally, substitute back u = cos(t):
= (1/3cos^3(t)) - (1/cos(t)) + C
Therefore, the integral of (sin^2(t)+1)/(cos^4(t)) dt is equal to (1/3cos^3(t)) - (1/cos(t)) + C.