Porisma

Review practice · Section 1 of 3

Questions 1–6 in this collection

A derivative unlocks integration by parts · Use a double-angle identity and an exponential substitution · +4 more

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  1. 1

    (i) Find ddx(1−e4x)1/2\dfrac{d}{dx}(1-e^{4x})^{1/2}. mark

    (ii) Hence find ∫e2xsin⁡−1(e2x) dx\displaystyle\int e^{2x}\sin^{-1}(e^{2x})\,dx. marks

    [2013/IJC/I/3]

    CJC · Review Problems · Question 1

  2. 2

    Find

    (i) ∫11+cos⁡2x dx\displaystyle\int\dfrac1{1+\cos2x}\,dx. marks

    (ii) ∫e2xex+2 dx\displaystyle\int\dfrac{e^{2x}}{e^x+2}\,dx, using the substitution u=ex+2u=e^x+2. marks

    [2013/MI/I/8(i),(iii)]

    CJC · Review Problems · Question 2

  3. 3

    (a) Find ∫8x2+14x2−1 dx\displaystyle\int\dfrac{8x^2+1}{4x^2-1}\,dx. marks

    (b) Find, using integration by parts, ∫e3xtan⁡−1(e−3x) dx\displaystyle\int e^{3x}\tan^{-1}(e^{-3x})\,dx. marks

    [2013/SRJC/I/2]

    CJC · Review Problems · Question 3

  4. 4

    Use the substitution y=1−2xy=1-2x to find ∫x2+11−2x dx\displaystyle\int\dfrac{x^2+1}{\sqrt{1-2x}}\,dx. marks

    [2013/HCI/I/9(a)]

    CJC · Review Problems · Question 4

  5. 5

    (i) Show that ∫x1−x2 dx=−1−x2+c\displaystyle\int\dfrac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}+c, where cc is an arbitrary constant. marks

    (ii) Hence find ∫(sin⁡−1x)x1−x2 dx\displaystyle\int(\sin^{-1}x)\dfrac{x}{\sqrt{1-x^2}}\,dx. marks

    [2016/JJC/Promo/9(b)]

    CJC · Review Problems · Question 5

  6. 6

    (a) Use the substitution u=x+1u=\sqrt{x+1} to find ∫x2x+1 dx\displaystyle\int\dfrac{x^2}{\sqrt{x+1}}\,dx. marks

    (b) Find ∫e5x(e5x−e)4 dx\displaystyle\int\dfrac{e^{5x}}{(e^{5x}-e)^4}\,dx. marks

    [2016/MJC/Promo/7(a),(b)]

    CJC · Review Problems · Question 6

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