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Concept 05 / 10Integral Calculus

Special Integrals: Quadratics, Square Roots and Trigonometric Denominators

Integrals of 1/(ax^2 + bx + c) and 1/sqrt(ax^2 + bx + c) by completing the square, linear-over-quadratic forms by splitting the numerator, 1/(a + b cos x) and 1/(a sin x + b cos x) types, and the derivative-of-denominator split for (a sin x + b cos x)/(c sin x + d cos x).

Completing the square is the master key for integrands involving a quadratic ax2+bx+cax^{2} + bx + c. Writing ax2+bx+c=a[(x+b2a)2+k]ax^{2} + bx + c = a[(x + \frac{b}{2a})^{2} + k] reduces every such integral to one of six standard forms (with u=x+b2au = x + \frac{b}{2a}): ∫duu2+a2=1atan⁡−1ua\int\frac{du}{u^{2} + a^{2}} = \frac{1}{a}\tan^{-1}\frac{u}{a}. ∫duu2−a2=12aln⁡∣u−au+a∣\int\frac{du}{u^{2} - a^{2}} = \frac{1}{2a}\ln\left|\frac{u - a}{u + a}\right|. ∫dua2−u2=12aln⁡∣a+ua−u∣\int\frac{du}{a^{2} - u^{2}} = \frac{1}{2a}\ln\left|\frac{a + u}{a - u}\right|. ∫dua2−u2=sin⁡−1ua\int\frac{du}{\sqrt{a^{2} - u^{2}}} = \sin^{-1}\frac{u}{a}. ∫duu2+a2=ln⁡∣u+u2+a2∣\int\frac{du}{\sqrt{u^{2} + a^{2}}} = \ln|u + \sqrt{u^{2} + a^{2}}|. ∫duu2−a2=ln⁡∣u+u2−a2∣\int\frac{du}{\sqrt{u^{2} - a^{2}}} = \ln|u + \sqrt{u^{2} - a^{2}}|. (Add +C+C to each.) Which form arises depends on the sign of aa and of the constant left after completing the square. Linear over quadratic: ∫px+qax2+bx+cdx\int\frac{px + q}{ax^{2} + bx + c}dx. Split the numerator as a multiple of the derivative of the denominator plus a constant: px+q=λ(2ax+b)+μpx + q = \lambda(2ax + b) + \mu. The first part integrates to λln⁡∣ax2+bx+c∣\lambda\ln|ax^{2} + bx + c| and the second is one of the standard forms above. The same split works for ∫px+qax2+bx+cdx\int\frac{px + q}{\sqrt{ax^{2} + bx + c}}dx, where the first part gives 2λax2+bx+c2\lambda\sqrt{ax^{2} + bx + c}, and for ∫(px+q)ax2+bx+c dx\int(px + q)\sqrt{ax^{2} + bx + c}\,dx. Trigonometric denominators. ∫dxa+bcos⁡x\int\frac{dx}{a + b\cos x}, ∫dxa+bsin⁡x\int\frac{dx}{a + b\sin x} and ∫dxa+bsin⁡x+ccos⁡x\int\frac{dx}{a + b\sin x + c\cos x}: use t=tan⁡x2t = \tan\frac{x}{2}, which turns them into integrals of 1quadratic in t\frac{1}{\text{quadratic in } t}. For ∫dxa+bcos⁡x\int\frac{dx}{a + b\cos x} with a>∣b∣a > |b| the result is 2a2−b2tan⁡−1(a−ba+btan⁡x2)+C\frac{2}{\sqrt{a^{2} - b^{2}}}\tan^{-1}\left(\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right) + C. ∫dxasin⁡x+bcos⁡x\int\frac{dx}{a\sin x + b\cos x}: write asin⁡x+bcos⁡x=rsin⁡(x+α)a\sin x + b\cos x = r\sin(x + \alpha) with r=a2+b2r = \sqrt{a^{2} + b^{2}}; the integral becomes 1r∫csc⁡(x+α)dx=1rln⁡∣tan⁡x+α2∣+C\frac{1}{r}\int\csc(x + \alpha)dx = \frac{1}{r}\ln|\tan\frac{x + \alpha}{2}| + C. ∫dxa+bsin⁡2x\int\frac{dx}{a + b\sin^{2}x} or ∫dxasin⁡2x+bcos⁡2x\int\frac{dx}{a\sin^{2}x + b\cos^{2}x}: divide the numerator and denominator by cos⁡2x\cos^{2}x and substitute u=tan⁡xu = \tan x. ∫asin⁡x+bcos⁡xcsin⁡x+dcos⁡xdx\int\frac{a\sin x + b\cos x}{c\sin x + d\cos x}dx: write the numerator as λ(denominator)+μ(derivative of denominator)\lambda(\text{denominator}) + \mu(\text{derivative of denominator}). The integral is then λx+μln⁡∣csin⁡x+dcos⁡x∣+C\lambda x + \mu\ln|c\sin x + d\cos x| + C. Find λ\lambda and μ\mu by comparing the coefficients of sin⁡x\sin x and cos⁡x\cos x. Irrational forms. For ∫dx(px+q)ax+b\int\frac{dx}{(px + q)\sqrt{ax + b}} put t2=ax+bt^{2} = ax + b. For ∫dx(px+q)ax2+bx+c\int\frac{dx}{(px + q)\sqrt{ax^{2} + bx + c}} put px+q=1tpx + q = \frac{1}{t}. For ∫dx(x2+a)x2+b\int\frac{dx}{(x^{2} + a)\sqrt{x^{2} + b}} put x=1tx = \frac{1}{t} and then u2=1+bt2u^{2} = 1 + bt^{2} (or similar). Key Takeaways: - Complete the square; then match one of the six standard forms. - px+q=λ(derivative of quadratic)+μpx + q = \lambda(\text{derivative of quadratic}) + \mu. - ∫dxu2±a2=ln⁡∣u+u2±a2∣\int\frac{dx}{\sqrt{u^{2} \pm a^{2}}} = \ln|u + \sqrt{u^{2} \pm a^{2}}|. - t=tan⁡x2t = \tan\frac{x}{2} for 1a+bcos⁡x+csin⁡x\frac{1}{a + b\cos x + c\sin x}. - 1asin⁡2x+bcos⁡2x\frac{1}{a\sin^{2}x + b\cos^{2}x}: divide by cos⁡2x\cos^{2}x, put u=tan⁡xu = \tan x. - asin⁡x+bcos⁡xcsin⁡x+dcos⁡x\frac{a\sin x + b\cos x}{c\sin x + d\cos x}: numerator =λD+μD′= \lambda D + \mu D'. - Pitfall: a quadratic with negative leading coefficient must be written as −(…)-(\ldots) before completing the square, which changes the standard form used. Brain-Teaser 1: Evaluate ∫dx2x−x2\int\frac{dx}{\sqrt{2x - x^{2}}} and ∫2x+3x2+2x+5dx\int\frac{2x + 3}{x^{2} + 2x + 5}dx. Solution: First: 2x−x2=1−(x−1)22x - x^{2} = 1 - (x - 1)^{2}, so the integral is sin⁡−1(x−1)+C\sin^{-1}(x - 1) + C. Second: 2x+3=(2x+2)+12x + 3 = (2x + 2) + 1, and x2+2x+5=(x+1)2+4x^{2} + 2x + 5 = (x + 1)^{2} + 4. So the integral is ln⁡(x2+2x+5)+∫dx(x+1)2+4=ln⁡(x2+2x+5)+12tan⁡−1x+12+C\ln(x^{2} + 2x + 5) + \int\frac{dx}{(x + 1)^{2} + 4} = \ln(x^{2} + 2x + 5) + \frac{1}{2}\tan^{-1}\frac{x + 1}{2} + C. (No absolute value is needed, since the quadratic is always positive.) Brain-Teaser 2: Evaluate ∫3sin⁡x+2cos⁡x3cos⁡x+2sin⁡xdx\int\frac{3\sin x + 2\cos x}{3\cos x + 2\sin x}dx. Solution: Let D=2sin⁡x+3cos⁡xD = 2\sin x + 3\cos x, so D′=2cos⁡x−3sin⁡xD' = 2\cos x - 3\sin x. Write 3sin⁡x+2cos⁡x=λ(2sin⁡x+3cos⁡x)+μ(2cos⁡x−3sin⁡x)3\sin x + 2\cos x = \lambda(2\sin x + 3\cos x) + \mu(2\cos x - 3\sin x). Comparing: sin⁡x\sin x: 2λ−3μ=32\lambda - 3\mu = 3; cos⁡x\cos x: 3λ+2μ=23\lambda + 2\mu = 2. Solving: λ=1213\lambda = \frac{12}{13}, μ=−513\mu = -\frac{5}{13}. So the integral is 1213x−513ln⁡∣2sin⁡x+3cos⁡x∣+C\frac{12}{13}x - \frac{5}{13}\ln|2\sin x + 3\cos x| + C. The split into 'λD+μD′\lambda D + \mu D'' always works for linear combinations of sin⁡\sin and cos⁡\cos, because DD and D′D' span the same two-dimensional space. Brain-Teaser 3: Evaluate ∫dx5+4cos⁡x\int\frac{dx}{5 + 4\cos x}. Solution: With t=tan⁡x2t = \tan\frac{x}{2}: 5+4cos⁡x=5+4(1−t2)1+t2=9+t21+t25 + 4\cos x = 5 + \frac{4(1 - t^{2})}{1 + t^{2}} = \frac{9 + t^{2}}{1 + t^{2}}, and dx=2dt1+t2dx = \frac{2dt}{1 + t^{2}}. The integral becomes ∫2dtt2+9=23tan⁡−1t3+C=23tan⁡−1(13tan⁡x2)+C\int\frac{2dt}{t^{2} + 9} = \frac{2}{3}\tan^{-1}\frac{t}{3} + C = \frac{2}{3}\tan^{-1}\left(\frac{1}{3}\tan\frac{x}{2}\right) + C. This matches the general formula with a=5a = 5, b=4b = 4: 23tan⁡−1(19tan⁡x2)\frac{2}{3}\tan^{-1}(\sqrt{\frac{1}{9}}\tan\frac{x}{2}).

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Special Integrals: Quadratics, Square Roots and Trigonometric Denominators — DPP

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