How do you integrate using u-substitution?

Question

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

Answer 1

Favorite 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

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