Techniques of Integration: math test solutions, College – download the PDF

1 Integration by parts / 4 pts

(a) \(u=\ln x\), \(dv=x^2dx\), \(v=\tfrac{x^3}3\) (1 pt). \( \int=\tfrac{x^3}3\ln x-\tfrac13\int x^2dx=\tfrac{x^3}3\ln x-\tfrac{x^3}9+C \) (1 pt).

(b) \(u=x\), \(dv=\sin x\,dx\): \( [-x\cos x]_0^{\pi}+\int_0^{\pi}\cos x\,dx \) (1 pt) \(=\pi+0=\pi\) (1 pt).

2 Trigonometric integrals / 3 pts

(a) \( \cos^2x=\tfrac{1+\cos2x}2 \) so \( \int=\tfrac x2+\tfrac{\sin2x}4+C \) (1 pt).

(b) \( \cos^3x=(1-\sin^2x)\cos x \) (1 pt); with \(u=\sin x\): \( \int(u^2-u^4)du=\tfrac{\sin^3x}3-\tfrac{\sin^5x}5+C \) (1 pt).

3 Trigonometric substitution / 4 pts

(a) \(x=2\sin\theta\), \(dx=2\cos\theta\,d\theta\), \( \sqrt{4-x^2}=2\cos\theta \) (1 pt). The integrand becomes \( \dfrac1{4\sin^2\theta} \) (1 pt) and \( \int=-\tfrac14\cot\theta+C=-\dfrac{\sqrt{4-x^2}}{4x}+C \) (1 pt).

(b) \( \Big[\tfrac13\arctan\tfrac x3\Big]_0^3=\tfrac13\cdot\tfrac\pi4=\tfrac\pi{12} \) (1 pt).

4 Partial fractions / 4 pts

(a) \(x^2-x-6=(x-3)(x+2)\); \(A=\tfrac85\), \(B=-\tfrac35\); so \( f=\dfrac{8/5}{x-3}-\dfrac{3/5}{x+2} \) (2 pts).

(b) \( \tfrac85\ln|x-3|-\tfrac35\ln|x+2|+C \) (1 pt).

(c) \( \tfrac85\ln3-\tfrac35\ln\tfrac86=\tfrac85\ln3-\tfrac35\ln\tfrac43\approx1.757780-0.172609\approx1.585 \) (1 pt).

5 Improper integrals / 3 pts

(a) By parts: \( \int\dfrac{\ln x}{x^2}dx=-\dfrac{\ln x}x-\dfrac1x \) (1 pt). At \(\infty\) it tends to 0, at 1 it is \(-1\); the integral equals \(1\) (1 pt).

(b) \( \Big[\tfrac32x^{2/3}\Big]_0^1=\tfrac32 \) (1 pt).

6 Numerical integration / 2 pts

\(h=0.5\); values \(0,\ 0.25,\ 1,\ 2.25,\ 4\). \( T_4=0.25\big[0+2(0.25+1+2.25)+4\big]=2.75 \) (1 pt). The exact value is \( \tfrac83\approx2.667 \), so the error is \( \tfrac1{12}\approx0.083 \) (1 pt).