MathHawk
May28-09, 11:40 PM
I like to work these problems out and then check them with online integral calculators.
1. The problem statement, all variables and given/known data
\int^{\pi/4}_{0}tan2x * sec4x dx
2. Relevant equations
\frac{d}{dx}tanx = sec2x
sec2x = 1 + tan2x
3. The attempt at a solution
This seems so simple, using the identities and u substitution:
\int^{\pi/4}_{0}tan2x * sec4x dx
\int^{\pi/4}_{0}tan2x * sec2x * sec2x dx
\int^{\pi/4}_{0}tan2x * (tan2x + 1) * sec2x dx
\int^{\pi/4}_{0}(tan4x + tan2x) * sec2x dx
Now: u = tanx, du = sec2x dx. tan \pi/4 = 1, tan 0 = 0/
\int^{1}_{0}(u4 + u2) du
= [\frac{1}{5}u5 + \frac{1}{3}u3]^{1}_{0}
\frac{1}{5} + \frac{1}{3} - (0 + 0) = \frac{8}{15}
Therefore:
\int^{\pi/4}_{0}tan2x * sec4x dx = \frac{8}{15}
Online indefinite integral calculators disagree with my indefinite integral, and the definite integral calculator I tried timed out. This looks flawless to me, but apparently I'm an idiot.
Thank you very much in advance for any help.
1. The problem statement, all variables and given/known data
\int^{\pi/4}_{0}tan2x * sec4x dx
2. Relevant equations
\frac{d}{dx}tanx = sec2x
sec2x = 1 + tan2x
3. The attempt at a solution
This seems so simple, using the identities and u substitution:
\int^{\pi/4}_{0}tan2x * sec4x dx
\int^{\pi/4}_{0}tan2x * sec2x * sec2x dx
\int^{\pi/4}_{0}tan2x * (tan2x + 1) * sec2x dx
\int^{\pi/4}_{0}(tan4x + tan2x) * sec2x dx
Now: u = tanx, du = sec2x dx. tan \pi/4 = 1, tan 0 = 0/
\int^{1}_{0}(u4 + u2) du
= [\frac{1}{5}u5 + \frac{1}{3}u3]^{1}_{0}
\frac{1}{5} + \frac{1}{3} - (0 + 0) = \frac{8}{15}
Therefore:
\int^{\pi/4}_{0}tan2x * sec4x dx = \frac{8}{15}
Online indefinite integral calculators disagree with my indefinite integral, and the definite integral calculator I tried timed out. This looks flawless to me, but apparently I'm an idiot.
Thank you very much in advance for any help.