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integral of (tanx)^4
(tanx)^4 = (tanx)^2 (tanx)^2
=(sec^2 x - 1)(tan^2 x)
=(sec^2 x)(tan^2 x) - tan^2 x
= integral of sec^2 x tan^2 x dx - integral of tan^2 x dx
First, integral of sec^2 x tan^2 x dx
Let u = tanx because that would make du = sec^2 x dx
so then we have integral of u^2 du which is (1/3)u^3
substituting back in tanx we get (1/3)tan^3 x
Next, integral of tan^2 x
tan^2 x = sec^2 x -1
integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx
= tanx - x
so putting it all together we have integral of tan^4 x dx =
(1/3)tan^3 x - tanx + x + C
Tiffany Wart
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