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Concept 02 / 10Integral Calculus

Integration by Substitution and Standard Trigonometric Substitutions

Changing the variable with u = g(x) and du = g'(x)dx, recognising f(g(x))g'(x), the forms f'(x)/f(x) and f'(x)[f(x)]^n, trigonometric substitutions for square roots of quadratics, and integrals of powers of sine and cosine.

Substitution is the chain rule in reverse. Since ddxF(g(x))=F′(g(x))g′(x)\frac{d}{dx}F(g(x)) = F'(g(x))g'(x), ∫f(g(x))g′(x) dx=∫f(u) du\int f(g(x))g'(x)\,dx = \int f(u)\,du with u=g(x)u = g(x), du=g′(x) dxdu = g'(x)\,dx. The signal to substitute is a composite function whose inner function's derivative also appears as a factor (up to a constant). After integrating in uu, replace uu by g(x)g(x). Two patterns are so common they should be recognised instantly. ∫f′(x)f(x)dx=ln⁡∣f(x)∣+C\int\frac{f'(x)}{f(x)}dx = \ln|f(x)| + C. For example, ∫2x+3x2+3x+7dx=ln⁡∣x2+3x+7∣+C\int\frac{2x + 3}{x^{2} + 3x + 7}dx = \ln|x^{2} + 3x + 7| + C, and ∫tan⁡x dx=∫sin⁡xcos⁡xdx=−ln⁡∣cos⁡x∣+C\int\tan x\,dx = \int\frac{\sin x}{\cos x}dx = -\ln|\cos x| + C. ∫f′(x)[f(x)]ndx=[f(x)]n+1n+1+C\int f'(x)[f(x)]^{n}dx = \frac{[f(x)]^{n+1}}{n + 1} + C for n≠−1n \neq -1. For example, ∫sin⁡3xcos⁡x dx=sin⁡4x4+C\int\sin^{3}x\cos x\,dx = \frac{\sin^{4}x}{4} + C. Also ∫f′(x)ef(x)dx=ef(x)+C\int f'(x)e^{f(x)}dx = e^{f(x)} + C. Choosing the substitution. Try the inner function of the most complicated composite, or the expression under a root, or the exponent. If the derivative of the chosen uu is not present, the substitution may still work after rearranging, for example writing the remaining xx in terms of uu: for ∫xx+1 dx\int x\sqrt{x + 1}\,dx, put u=x+1u = x + 1, x=u−1x = u - 1, giving ∫(u−1)u du\int(u - 1)\sqrt{u}\,du. Trigonometric substitutions for roots of quadratics. a2−x2\sqrt{a^{2} - x^{2}}: x=asin⁡θx = a\sin\theta, since a2−a2sin⁡2θ=a2cos⁡2θa^{2} - a^{2}\sin^{2}\theta = a^{2}\cos^{2}\theta. a2+x2\sqrt{a^{2} + x^{2}} or a2+x2a^{2} + x^{2}: x=atan⁡θx = a\tan\theta. x2−a2\sqrt{x^{2} - a^{2}}: x=asec⁡θx = a\sec\theta. a−xa+x\sqrt{\frac{a - x}{a + x}}: x=acos⁡2θx = a\cos 2\theta. (x−α)(β−x)\sqrt{(x - \alpha)(\beta - x)}: x=αcos⁡2θ+βsin⁡2θx = \alpha\cos^{2}\theta + \beta\sin^{2}\theta. After integrating in θ\theta, convert back using a right triangle built from the substitution. Powers of sine and cosine. For ∫sin⁡mxcos⁡nx dx\int\sin^{m}x\cos^{n}x\,dx: If one exponent is odd, save one factor of that function, convert the rest using sin⁡2+cos⁡2=1\sin^{2} + \cos^{2} = 1, and substitute the other function. For ∫sin⁡3x dx=∫(1−cos⁡2x)sin⁡x dx\int\sin^{3}x\,dx = \int(1 - \cos^{2}x)\sin x\,dx, put u=cos⁡xu = \cos x: −cos⁡x+cos⁡3x3+C-\cos x + \frac{\cos^{3}x}{3} + C. If both are even, use the half-angle formulas sin⁡2x=1−cos⁡2x2\sin^{2}x = \frac{1 - \cos 2x}{2} and cos⁡2x=1+cos⁡2x2\cos^{2}x = \frac{1 + \cos 2x}{2} repeatedly. If m+nm + n is a negative even integer, divide by cos⁡m+nx\cos^{m+n}x and substitute u=tan⁡xu = \tan x. Other useful substitutions: u=exu = e^{x} for rational functions of exe^{x}; u=x1/ku = x^{1/k} where kk is the LCM of the root orders for integrands with several roots of xx; u=1xu = \frac{1}{x} for ∫dxxn…\int\frac{dx}{x^{n}\sqrt{\ldots}} types; and the tangent half-angle t=tan⁡x2t = \tan\frac{x}{2}, with sin⁡x=2t1+t2\sin x = \frac{2t}{1 + t^{2}}, cos⁡x=1−t21+t2\cos x = \frac{1 - t^{2}}{1 + t^{2}}, dx=2dt1+t2dx = \frac{2dt}{1 + t^{2}}, which turns any rational function of sin⁡x\sin x and cos⁡x\cos x into a rational function of tt. Key Takeaways: - ∫f(g(x))g′(x)dx=∫f(u)du\int f(g(x))g'(x)dx = \int f(u)du with u=g(x)u = g(x). - ∫f′f=ln⁡∣f∣\int\frac{f'}{f} = \ln|f|; ∫f′fn=fn+1n+1\int f'f^{n} = \frac{f^{n+1}}{n + 1}; ∫f′ef=ef\int f'e^{f} = e^{f}. - a2−x2\sqrt{a^{2} - x^{2}}: x=asin⁡θx = a\sin\theta; a2+x2a^{2} + x^{2}: x=atan⁡θx = a\tan\theta; x2−a2\sqrt{x^{2} - a^{2}}: x=asec⁡θx = a\sec\theta. - Odd power of sin⁡\sin or cos⁡\cos: save one factor and substitute the other function. - Even powers: half-angle formulas. - t=tan⁡x2t = \tan\frac{x}{2} rationalises any R(sin⁡x,cos⁡x)R(\sin x, \cos x). - Pitfall: forgetting to convert dxdx into dudu, or leaving the answer in terms of uu. Brain-Teaser 1: Evaluate ∫dxx(x5+1)\int\frac{dx}{x(x^{5} + 1)}. Solution: Multiply the numerator and denominator by x4x^{4}: ∫x4dxx5(x5+1)\int\frac{x^{4}dx}{x^{5}(x^{5} + 1)}. Put u=x5u = x^{5}, du=5x4dxdu = 5x^{4}dx: 15∫duu(u+1)=15∫(1u−1u+1)du=15ln⁡∣x5x5+1∣+C\frac{1}{5}\int\frac{du}{u(u + 1)} = \frac{1}{5}\int\left(\frac{1}{u} - \frac{1}{u + 1}\right)du = \frac{1}{5}\ln\left|\frac{x^{5}}{x^{5} + 1}\right| + C. Multiplying by a suitable power of xx to create the derivative of the inner function is a standard trick for ∫dxx(xn+1)\int\frac{dx}{x(x^{n} + 1)}. Brain-Teaser 2: Evaluate ∫sin⁡2xsin⁡4x+cos⁡4xdx\int\frac{\sin 2x}{\sin^{4}x + \cos^{4}x}dx. Solution: Divide the numerator and denominator by cos⁡4x\cos^{4}x: 2tan⁡xsec⁡2xtan⁡4x+1\frac{2\tan x\sec^{2}x}{\tan^{4}x + 1}. Put u=tan⁡2xu = \tan^{2}x, du=2tan⁡xsec⁡2x dxdu = 2\tan x\sec^{2}x\,dx: ∫duu2+1=tan⁡−1u+C=tan⁡−1(tan⁡2x)+C\int\frac{du}{u^{2} + 1} = \tan^{-1}u + C = \tan^{-1}(\tan^{2}x) + C. The denominator of degree 44 in sin⁡\sin and cos⁡\cos suggests dividing by cos⁡4x\cos^{4}x; the numerator then becomes exactly the derivative of tan⁡2x\tan^{2}x. Brain-Teaser 3: Evaluate ∫1−x1+x dx\int\sqrt{\frac{1 - x}{1 + x}}\,dx. Solution: Put x=cos⁡2θx = \cos 2\theta, dx=−2sin⁡2θ dθdx = -2\sin 2\theta\,d\theta. Then 1−cos⁡2θ1+cos⁡2θ=tan⁡θ\sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} = \tan\theta (for 0<θ<π20 < \theta < \frac{\pi}{2}). The integral becomes −∫tan⁡θ⋅2sin⁡2θ dθ=−∫4sin⁡2θ dθ=−∫2(1−cos⁡2θ)dθ=−2θ+sin⁡2θ+C-\int\tan\theta \cdot 2\sin 2\theta\,d\theta = -\int 4\sin^{2}\theta\,d\theta = -\int 2(1 - \cos 2\theta)d\theta = -2\theta + \sin 2\theta + C. Back-substitute 2θ=cos⁡−1x2\theta = \cos^{-1}x and sin⁡2θ=1−x2\sin 2\theta = \sqrt{1 - x^{2}}: the answer is 1−x2−cos⁡−1x+C\sqrt{1 - x^{2}} - \cos^{-1}x + C. Differentiating confirms it: −x1−x2+11−x2=1−x1−x2=1−x1+x\frac{-x}{\sqrt{1 - x^{2}}} + \frac{1}{\sqrt{1 - x^{2}}} = \frac{1 - x}{\sqrt{1 - x^{2}}} = \sqrt{\frac{1 - x}{1 + x}}.

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Integration by Substitution and Standard Trigonometric Substitutions — DPP

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