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I like to work these problems out and then check them with online integral calculators.

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

[tex]\frac{d}{dx}[/tex]tanx = sec

sec

This seems so simple, using the identities and u substitution:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex](tan

Now: u = tanx, du = sec

[tex]\int[/tex][tex]^{1}_{0}[/tex](u

= [[tex]\frac{1}{5}[/tex]u

[tex]\frac{1}{5}[/tex] + [tex]\frac{1}{3}[/tex] - (0 + 0) = [tex]\frac{8}{15}[/tex]

Therefore:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

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.

## Homework Statement

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

^{2}x * sec^{4}x dx## Homework Equations

[tex]\frac{d}{dx}[/tex]tanx = sec

^{2}xsec

^{2}x = 1 + tan^{2}x## The Attempt at a Solution

This seems so simple, using the identities and u substitution:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

^{2}x * sec^{4}x dx[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

^{2}x * sec^{2}x * sec^{2}x dx[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

^{2}x * (tan^{2}x + 1) * sec^{2}x dx[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex](tan

^{4}x + tan^{2}x) * sec^{2}x dxNow: u = tanx, du = sec

^{2}x dx. tan [tex]\pi/4[/tex] = 1, tan 0 = 0/[tex]\int[/tex][tex]^{1}_{0}[/tex](u

^{4}+ u^{2}) du= [[tex]\frac{1}{5}[/tex]u

^{5}+ [tex]\frac{1}{3}[/tex]u^{3}][tex]^{1}_{0}[/tex][tex]\frac{1}{5}[/tex] + [tex]\frac{1}{3}[/tex] - (0 + 0) = [tex]\frac{8}{15}[/tex]

Therefore:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan

^{2}x * sec^{4}x dx = [tex]\frac{8}{15}[/tex]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.

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