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Discussion questions · Section 2 of 2

Questions 7–12 in this collection

Use the specified substitutions · Integration by parts: logarithms, powers and a returning integral · +4 more

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  1. 7

    Use the given substitutions to find the following integrals:

    (a) ∫1ex+e−x dx\displaystyle\int\dfrac1{e^x+e^{-x}}\,dx, let u=exu=e^x.

    (b) ∫x+1(2x−1)2 dx\displaystyle\int\dfrac{x+1}{(2x-1)^2}\,dx, let u=2x−1u=2x-1.

    (c) ∫12+x dx\displaystyle\int\dfrac1{2+\sqrt x}\,dx, let u=xu=\sqrt x.

    (d) ∫sin⁡3x dx\displaystyle\int\sin^3x\,dx, let u=cos⁡xu=\cos x.

    (e) ∫x1−x2 dx\displaystyle\int\dfrac{x}{\sqrt{1-x^2}}\,dx, let x=sin⁡θx=\sin\theta.

    (f) ∫1(x2+1)2 dx\displaystyle\int\dfrac1{(x^2+1)^2}\,dx, let x=tan⁡θx=\tan\theta.

    CJC · Discussion · Question 7

  2. 8

    Find the following integrals:

    (a) ∫x2ln⁡x dx\displaystyle\int x^2\ln x\,dx

    (b) ∫ln⁡(x2+1) dx\displaystyle\int \ln(x^2+1)\,dx

    (c) ∫xtan⁡−1(x2) dx\displaystyle\int x\tan^{-1}(x^2)\,dx

    (d) ∫x2ex dx\displaystyle\int x^2e^x\,dx

    (e) ∫excos⁡x dx\displaystyle\int e^x\cos x\,dx

    (f) ∫θsin⁡θcos⁡θ dθ\displaystyle\int \theta\sin\theta\cos\theta\,d\theta.

    CJC · Discussion · Question 8

  3. 9

    (i) Show that ddx1−x2=−x1−x2\dfrac d{dx}\sqrt{1-x^2}=\dfrac{-x}{\sqrt{1-x^2}}, ∣x∣<1|x|<1.

    (ii) Hence find ∫sin⁡−1x dx\displaystyle\int\sin^{-1}x\,dx.

    CJC · Discussion · Question 9

  4. 10

    (i) State the derivative of sin⁡(x2)\sin(x^2).

    (ii) Find ∫x3cos⁡(x2) dx\displaystyle\int x^3\cos(x^2)\,dx.

    CJC · Discussion · Question 10

  5. 11

    You are given that I=∫x(1−x)1/2 dxI=\displaystyle\int x(1-x)^{1/2}\,dx.

    (i) Use integration by parts to find an expression for II.

    (ii) Use the substitution u2=1−xu^2=1-x to find another expression for II.

    (iii) Show algebraically that your answers to parts (i) and (ii) differ by a constant.

    [N2019/II/1]

    CJC · Discussion · Question 11

  6. 12

    Given 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B=\sin(A+B)-\sin(A-B), find ∫cos⁡4θsin⁡2θ dθ\displaystyle\int\cos4\theta\sin2\theta\,d\theta.

    CJC · Discussion · Question 12

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