Evaluate the integral?

Question

∫ Xtan^2(x^2)sec^2(x^2) dx


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Answer 1

Let u = x ²
du / 2 = x dx
I = (1/2) ∫ (tan ² u) (sec ² u) du
let b = tan u
db = sec ² u du
I = (1/2) ∫ b ² db
I = (1/2) (b ³ / 3) + C
I = (1/6) tan ³ u + C
I = (1/6) tan ³ (x ²) + C

Answer 2

First, note that (tan x)' = sec^2 x. So, it follows from the chain rule that (tan x^2)' = sec(x^2) 2x. Hence, your integral can be written as

(1/2) Integral tan^2(x^2) sec^2(x^2) (2x) dx =
= (1/2) Integral tan^2(x^2) (tan(x^2)' dx

Now it's easy , we have the square of a function multiplied by the derivative of the function. So,

(1/2) Integral tan^2(x^2) (tan(x^2)' dx =(1/2) (tan^3(x^2))/3 + C = (tan^3(x^2))/6 + C.

Answer 3

let, x^2 = u, du = 2xdx

we have:

(1/2)∫ tan^2(u)sec^2(u) du
= (1/2)∫ [ {sin^2(u)/cos^2(u)} / {1/(cos^2(u))} ] du
= (1/2)∫ sin^2(u) du
= (1/2)∫ (1/2){1-cos(2u)} du [since: sin^2(X) = (1/2)(1-cos2X)]
= (1/4)∫ {1-cos(2u)} du
= (1/4) [ u - (1/2)sin(2u) ] + C
= (1/4) [ x^2 - (1/2)sin(2x^2) ] + C

for an arbitrary constant, C