How did this person get the power of u^-2/3? integral_-1^1 3 Squareroot sec(x).

Question

How did this person get the power of u^-2/3? integral_-1^1 3 Squareroot sec(x). Tan(x) dx = integral 3 Squareroot sec(x). Tan(x) dx Apply integral substitution u = sec(x): du = u Tan(x) dx. = integral 1/u^2/3 du = integral u^-2/3 du. = u^-2/3+1/-2/3+1 = (sec(x))^-2/3+1/-2/3+1 = 3 3 Squareroot sec(x) + C integral_-1^1 3 Squareroot sec(x) middot Tan(x) dx = [3 3 Squareroot sec(x)]_1^1 Therefore for an add function f(x) = -f(-x): integral_a^a f(x) dx = 0 Therefore integral_-1^1 3 Squareroot sec(x) middot Tan(x) dx = 0

Explanation / Answer

Solution:

This is right;

sec(x) * tan(x) dx

Apply integral substitution;

u = sec(x) => du = sec(x) tan(x) dx => du = u tan(x) dx => tan(x) dx = (1/u) du

u * (1/u) du

= (u)^(1/3) * (1/u) du

= (u)^(1/3) * (u)^(-1) du

= (u)^(1/3 - 1) du

= (u)^(-2/3) du

This is the complete process to get u^(-2/3).

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