How do you integrate using u-substitution?
Does anyone know how to evaluate these using u-substitution??
∫ e^x sec² (e^x) dx
∫ sin(√x) / √x dx
∫ sin(theta) cos(theta) / sin²(theta)+1 d(theta)
thank you, i will choose best answer for the quickest correct answer, or whoever answers the most correct
1 Answer
- germanoLv 79 years agoBest Answer
Hello,
∫ e^x sec²(e^x) dx =
rewrite it as:
∫ sec²(e^x) e^x dx =
let:
e^x = u
differentiate both sides:
d(e^x) = du
e^x dx = du
then, substituting:
∫ sec²(e^x) e^x dx = ∫ sec²u du =
∫ d(tan u) =
tan u + C
being u = e^x, the answer is:
∫ e^x sec²(e^x) dx = tan(e^x) + C
======================================
∫ [sin(√x) /√x] dx =
rewrite it as:
∫ sin(√x) (1/√x) dx =
let:
√x = u
differentiate both sides:
d(√x) = d[x^(1/2)] = du
(1/2) x^[(1/2) -1] dx = du
(1/2)x^(-1/2) dx = du
(1/2) (1/√x) dx = du
(1/√x) dx = 2 du
thus, substituting:
∫ sin(√x) (1/√x) dx = ∫ sin u 2 du =
(pulling the constant out)
2 ∫ sin u du =
2 (- cos u) + C =
- 2cos u + C
substitute back √x for u, ending with:
∫ [sin(√x) /√x] dx = - 2cos(√x) + C
=====================================
∫ [sinθ cosθ /(sin²θ + 1)] dθ =
rewrite it as:
∫ [1 /(sin²θ + 1)] sinθ cosθ dθ =
let:
(sin²θ + 1) = u
differentiate both sides:
d(sin²θ + 1) = du
2sinθ cosθ dθ = du
sinθ cosθ dθ = (1/2) du
then, substituting:
∫ [1 /(sin²θ + 1)] sinθ cosθ dθ = ∫ (1 /u) (1/2) du =
(pulling the constant out)
(1/2) ∫ (1 /u) du =
(1/2) ln | u | + C
being u = sin²θ + 1, the answer is:
∫ [sinθ cosθ /(sin²θ + 1)] dθ = (1/2) ln (sin²θ + 1) + C
(the absolute value is not needed any longer since the argument (sin²θ + 1) is always positive)
I hope it's helpful