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Techniques of integration

Practice section 8 of 10

Questions 43–48 in this collection

Practice questions

Review practice · 2 of 6

Use a shown antiderivative inside integration by parts

(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. [2]

(ii) Hence find ∫(sin⁡−1x)x1−x2 dx\displaystyle\int(\sin^{-1}x)\dfrac{x}{\sqrt{1-x^2}}\,dx. [4]

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

CJC · Review Problems · Question 5

Solution

Review practice · 3 of 6

Transform a square-root integral and reverse an exponential chain

(a) Use the substitution u=x+1u=\sqrt{x+1} to find ∫x2x+1 dx\displaystyle\int\dfrac{x^2}{\sqrt{x+1}}\,dx. [4]

(b) Find ∫e5x(e5x−e)4 dx\displaystyle\int\dfrac{e^{5x}}{(e^{5x}-e)^4}\,dx. [3]

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

CJC · Review Problems · Question 6

Solution

Review practice · 4 of 6

Apply double-angle and product-to-sum identities before integrating

Find the following integrals:

(a) ∫11+cos⁡4θ dθ\displaystyle\int\dfrac1{1+\cos4\theta}\,d\theta, [2]

(b) ∫cos⁡2xsin⁡3x dx\displaystyle\int\cos2x\sin3x\,dx, given 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B). [3]

[2016/NYJC/Promo/3 modified]

CJC · Review Problems · Question 7

Solution

Review practice · 5 of 6

Complete a square after substitution and use a product derivative

(a) By using the substitution u=xu=\sqrt x, find ∫1x (x+2x+2) dx\displaystyle\int\dfrac1{\sqrt x\,(x+2\sqrt x+2)}\,dx. [4]

(b) Find ddx(xsin⁡x)\dfrac d{dx}(x\sin x). [1]

Hence find ∫(sin⁡x+xcos⁡x)ln⁡x dx\displaystyle\int(\sin x+x\cos x)\ln x\,dx. [3]

[2016/PJC/Promo/7]

CJC · Review Problems · Question 8

Solution

Review practice · 6 of 6

Use the prescribed cosine substitution in a difference of squares

Use the substitution u=cos⁡2θu=\cos2\theta to find ∫sin⁡2θ4−cos⁡22θ dθ\displaystyle\int\dfrac{\sin2\theta}{4-\cos^22\theta}\,d\theta. [3]

[2018/DHS/Promo/1(b)]

CJC · Review Problems · Question 9

Solution

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