# Thread: Help solving Trig integral

1. ## Help solving Trig integral

Hey all, I need help finishing this integration problem

$\displaystyle \int_0^\frac{\pi}{2} \!16sec^4(\frac{t}{2}) \, dt$
$\displaystyle 16\int_0^\frac{\pi}{2} \!sec^4(\frac{t}{2}) \, dt$
$\displaystyle x=\frac{t}{2}$
$\displaystyle 2dx=dt$
$\displaystyle 32\int_0^\frac{\pi}{2} \!sec^2(x)sec^2(x) \, dx$
$\displaystyle 32[tan^2(x)] \bigg|_0^\frac{\pi}{2}$

Is this correct so far?

2. Originally Posted by mmattson07
Hey all, I need help finishing this integration problem

$\displaystyle \int_0^\frac{\pi}{2} \!16sec^4(\frac{t}{2}) \, dt$
$\displaystyle 16\int_0^\frac{\pi}{2} \!sec^4(\frac{t}{2}) \, dt$
$\displaystyle x=\frac{t}{2}$
$\displaystyle 2dx=dt$
$\displaystyle 16\int_0^\frac{\pi}{2} \!sec^2(x)sec^2(x) \, dx$
reset limits of integration ...

$\displaystyle 32 \int_0^\frac{\pi}{4} \sec^2{x}\sec^2{x} \, dx$

$\displaystyle 32 \int_0^\frac{\pi}{4} (1 + \tan^2{x})\sec^2{x} \, dx$

$\displaystyle u = \tan{x}$

$\displaystyle du = \sec^2{x} \, dx$

$\displaystyle 32 \int_0^1 (1 + u^2) \, du$

finish ...

3. ahh k thx dude

When i get here what limits do i use...

$\displaystyle 32[tan(\frac{t}{2})+\frac{1}{3}tan^3(\frac{t}{2})]\bigg|_0^\frac{\pi}{2}$ ???

Edit: That evaulated at $\displaystyle \frac{\pi}{2}$ would be $\displaystyle \frac{128}{3}$ yeeeee