Integration Strategy — Question 9

Section 8-7 · Calculus pages 1557–1558

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Question 9

Choose the best method for each integral. Do not evaluate. ∫x2exdx,∫dxx2−4,∫9−x2dx,∫sin⁡2xdx.\int x^2e^x\,dx,\qquad \int\frac{dx}{x^2-4},\qquad \int\sqrt{9-x^2}\,dx,\qquad \int\sin^2x\,dx.

Question 9 – Solution

1. For ∫x2exdx\int x^2e^x\,dx, differentiating x2x^2 repeatedly makes it simpler, while integrating exe^x changes nothing. Use integration by parts twice.

2. For ∫dx/(x2−4)\int dx/(x^2-4), factor x2−4=(x−2)(x+2).x^2-4=(x-2)(x+2). The denominator has distinct linear factors, so use partial fractions.

3. For ∫9−x2dx\int\sqrt{9-x^2}\,dx, the form a2−x2\sqrt{a^2-x^2} suggests x=asin⁡θx=a\sin\theta. Use trigonometric substitution with x=3sin⁡θx=3\sin\theta.

4. For ∫sin⁡2xdx\int\sin^2x\,dx, both sine and cosine powers are even. Use the power-reduction identity sin⁡2x=1−cos⁡2x2.\sin^2x=\frac{1-\cos2x}{2}.

x2exintegration by parts1/(x2−4)partial fractions9−x2trigonometric substitutionsin⁡2xpower reduction\boxed{\begin{array}{c|c} x^2e^x&\text{integration by parts}\\ 1/(x^2-4)&\text{partial fractions}\\ \sqrt{9-x^2}&\text{trigonometric substitution}\\ \sin^2x&\text{power reduction} \end{array}}