How do you evaluate the definite integral #intsec^2(t/4) # from #[0, pi]#?

Question

Ezra Adams · Accepted Answer

#int_0^pi sec^2(t/4)dt = 4# units

This is a relatively simple integral because the antiderivative of #sec^2(t)# is #tan(t)# (the derivative of tangent is secant squared; we just work in reverse when integrating). The only issue is that #t/4# within the secant function. To fix it, we reverse the chain rule - we multiply the result by 4. So: #int_0^pi sec^2(t/4)dt = [4tan(t/4)]_0^pi# Evaluating this from #0# to #pi#: #=(4tan((pi)/4)-4tan((0)/4))# #=(4(1)-4(0))# #=4# And that's it.

Cameron Alexander · Answer

undefined To evaluate the definite integral $ \int_{0}^{\pi} \sec^2(\frac{t}{4}) \, dt $ over the interval $[0, \pi]$, you can use a substitution. Let $ u = \frac{t}{4} $, so $ du = \frac{1}{4} dt $. Then the integral becomes $ 4\int_{0}^{\pi/4} \sec^2(u) \, du $.

The integral of $ \sec^2(u) $ is $ \tan(u) $. So, integrating $ \sec^2(u) $ from $0$ to $\pi/4$ gives $ \tan(\pi/4) - \tan(0) = 1 - 0 = 1 $.

Thus, the value of the definite integral $ \int_{0}^{\pi} \sec^2(\frac{t}{4}) \, dt $ is $ 4 \times 1 = 4 $.