Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

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

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  • 9 years ago
    Best 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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