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Section 8.3 #4: $$\displaystyle\int \sin^3(3x) \mathrm{d}x$$
Solution: Recall the Pythagorean identity $$\cos^2(\theta) + \sin^2(\theta)=1.$$ We applying this rearranged to $\sin^2(\theta)=1-\cos^2(\theta)$ with $\theta=3x$ yields $$\sin^3(3x) = \sin(3x) \sin^2(3x) = \sin(3x) (1-\cos^2(3x)).$$ Therefore we may compute $$\begin{array}{ll} \displaystyle\int \sin^3(3x) \mathrm{d}x &= \displaystyle\int \sin(3x) \left( 1-\cos^2(3x) \right) \mathrm{d}x \\ &= \displaystyle\int \sin(3x) \mathrm{d}x - \displaystyle\int \sin(3x) \cos^2(3x) \mathrm{d}x \\ &\stackrel{u_1=3x,u_2=\cos(3x)}{=} -\dfrac{1}{3} \displaystyle\int \sin(u_1) \mathrm{d}u_1 + \dfrac{1}{3} \displaystyle\int u_2^2 \mathrm{d}u_2 \\ &= \dfrac{1}{3} \cos(u_1) - \dfrac{1}{9} u_2^3 + C \\ &= \dfrac{1}{3} \cos(3x) - \dfrac{1}{9} \cos^3(3x) + C. \end{array}$$ Section 8.3 #9: $$\displaystyle\int \cos^2(3x) \mathrm{d}x$$
Solution: Recall the "reduction" identity $$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}{2}.$$ Apply that identity with $\theta=3x$ to see $$\cos^2(3x) = \dfrac{1+\cos(6x)}{2}.$$ Therefore compute $$\begin{array}{ll} \displaystyle\int \cos^2(3x) \mathrm{d}x &= \displaystyle\int \dfrac{1+\cos(6x)}{2} \mathrm{d}x \\ &= \dfrac{1}{2} \displaystyle\int 1 \mathrm{d}x + \dfrac{1}{2} \displaystyle\int \cos(6x) \mathrm{d}x \\ &=\dfrac{x}{2} + \dfrac{1}{12} \sin(6x) + C. \end{array}$$

Section 8.4 #22: $$\displaystyle\int \dfrac{x^2}{\sqrt{36-x^2}} \mathrm{d}x$$
Solution: Recall that the Pythagorean identity may be rearranged to say $$\sqrt{1-\sin^2(\theta)} = \cos(\theta).$$ Also recall the "reduction" identity $$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2}.$$ Let $x=6\sin(\theta)$ so that $\mathrm{d}x = 6\cos(\theta) \mathrm{d}\theta$. Substituting this variable into the integral yields $$\begin{array}{ll} \displaystyle\int \dfrac{x^2}{36-x^2} \mathrm{d}x &= \displaystyle\int \dfrac{36\sin^2(\theta)}{\sqrt{36-36\sin^2(\theta)}} 6 \cos(\theta) \mathrm{d}\theta \\ &= \dfrac{36 \cdot 6}{6} \displaystyle\int \dfrac{ \sin^2(\theta) \cos(\theta)}{\sqrt{1-\sin^2(\theta)}} \mathrm{d}\theta \\ &= 36 \displaystyle\int \dfrac{\sin^2(\theta)\cos(\theta)}{\cos(\theta)} \mathrm{d}\theta \\ &= 36 \displaystyle\int \sin^2(\theta) \mathrm{d}\theta \\ &= \dfrac{36}{2} \displaystyle\int 1 \mathrm{d}\theta - \dfrac{36}{2} \displaystyle\int \cos(2\theta) \mathrm{d}\theta \\ &= 18 \theta - 9 \sin(2\theta) + C. \end{array}$$ Solving our assignment $x=6\sin(\theta)$ for $\theta$ yields $\theta = \arcsin \left( \dfrac{x}{6} \right)$. Recall the double angle identity $$\cos(2\psi)=2\sin(\psi)\cos(\psi).$$ Applying this shows us that $$-9\sin(2\theta) = -18\sin(\theta)\cos(\theta) = -18 \left( \dfrac{x}{6} \right) \cos(\theta) = -3 \cos(\theta).$$ io determine $\cos(\theta)$ we must draw a triangle:

To find the side labelled "$?$", we use the Pythagorean theorem: $$?^2 + x^2 = 6^2.$$ Solving for $?$ yields $$? = \sqrt{36-x^2}.$$ Therefore $\cos(\theta) = \dfrac{\sqrt{36-x^2}}{6}$. So, finally, we compute $$\displaystyle\int \dfrac{x^2}{\sqrt{36-x^2}} \mathrm{d}x = 18 \arcsin \left( \dfrac{x}{6} \right) - 3x \cos(\theta) = 18\arcsin \left( \dfrac{x}{6} \right) - \dfrac{x}{2} \sqrt{36-x^2} +C.$$

Section 8.5 #6: $$\displaystyle\int \dfrac{2}{9x^2-1} \mathrm{d}x$$
Solution: The denominator is a difference of squares (i.e. $a^2-b^2=(a-b)(a+b)$) and so it factors: $$9x^2-1 = (3x)^2-1 = (3x-1)(3x+1).$$ Therefore, we must apply partial fractions to $$\dfrac{2}{9x^2-1} = \dfrac{A}{3x-1} + \dfrac{B}{3x+1}.$$ Multiply by the common denominator to get $$2 = A(3x+1) + B(3x-1).$$ Combining like-terms on the right yields $$2 = (3A+3B)x + (A-B).$$ Noting that the "coefficient of $x$" on the left is zero, we equate coefficients to get the following system of two linear equations in two variables ($A$ and $B$): $$\left\{ \begin{array}{lll} 3A+3B &= 0 & (i) \\ A-B &= 2 & (ii). \end{array} \right.$$ By $(ii)$ we conclude that $A=2+B$. Plug this into $(i)$ to get $$3(2+B)+3B=0,$$ or equivalently, $$6B = -6,$$ i.e. $B = -1$, and hence $A=2+(-1) = 1$. This gives us the partial fractions decomposition $$\dfrac{2}{9x^2-1} = \dfrac{1}{3x-1} - \dfrac{1}{3x+1}.$$ Therefore we may compute $$\begin{array}{ll} \displaystyle\int \dfrac{2}{9x^2-1} \mathrm{d}x &= \displaystyle\int \dfrac{1}{3x-1} \mathrm{d}x - \displaystyle\int \dfrac{1}{3x+1} \mathrm{d}x \\ &\stackrel{u=3x-1,v=3x+1}{=} \dfrac{1}{3} \displaystyle\int \dfrac{1}{u} \mathrm{d}u - \dfrac{1}{3} \displaystyle\int \dfrac{1}{v} \mathrm{d}v \\ &= \dfrac{1}{3} \log(u) - \dfrac{1}{3} \log(v) + C \\ &= \dfrac{1}{3} \log(3x-1) - \dfrac{1}{3} \log(3x+1) + C. \end{array}$$

Section 8.5 #9: $$\displaystyle\int \dfrac{x^2+12x+12}{x^3-4x} \mathrm{d}x$$
Solution: We may factor this denominator as $$x^3-4x = x(x^2-4) = x(x-2)(x+2),$$ so the partial fractions decomposition involves three terms (one for each linear factor): $$\dfrac{x^2+12x+12}{x^3-4x} = \dfrac{A}{x} + \dfrac{B}{x-2} + \dfrac{C}{x+2}.$$ Multiplying by the common denominator $x(x-2)(x+2)$ yields $$x^2+12x+12 = A(x-2)(x+2) + Bx(x+2) + Cx(x-2)=A(x^2-4) + B(x^2+2x) + C(x^2-2x).$$ Combining like-terms yields $$x^2+12x+12 = (A+B+C)x^2 + (2B-2C)x + (-4A).$$ Equating coefficients yields the following system of three linear equations in three variables: $$\left\{ \begin{array}{lll} A+B+C &= 1 & (i) \\ 2B-2C &= 12 & (ii) \\ -4A &= 12 & (iii). \end{array} \right.$$ From $(iii)$, we may conclude that $A=-3$. From $(ii)$, we may conclude that $B-C=6$, and hence $B=6+C$. Plugging $A=-3$ and $B=6+C$ into $(i)$, we get $$-3+(6+C)+C=1,$$ from which we may conclude $C=-1$ and $B=6+(-1)=5$, yielding the partial fractions decomposition $$\dfrac{x^2+12x+12}{x^3-4x} = -\dfrac{3}{x} + \dfrac{5}{x-2} - \dfrac{1}{x+2}.$$ Therefore we may compute $$\begin{array}{ll} \displaystyle\int \dfrac{x^2+12x+12}{x^3-4x} \mathrm{d}x &= -\displaystyle\int \dfrac{3}{x} \mathrm{d}x + 5 \displaystyle\int \dfrac{1}{x-2} \mathrm{d}x - \displaystyle\int \dfrac{1}{x+2} \mathrm{d}x \\ &= -3\log(x) + 5 \log(x-2) - \log(x+2) + C. \end{array}$$

Section 8.5 #13: $$\displaystyle\int \dfrac{4x^2+2x-1}{x^3+x^2} \mathrm{d}x$$
Solution: Factoring the denominator yields $$x^3+x^2 = x^2(x+1).$$ Therefore we set up the partial fraction equation: $$\dfrac{4x^2+2x-1}{x^3+x^2} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1}.$$ This yields a system of three equations in three variables and hence the following partial fractions decomposition: $$\dfrac{4x^2+2x-1}{x^3+x^2} = -\dfrac{1}{x^2} + \dfrac{1}{x+1} + \dfrac{3}{x}.$$ Therefore we compute $$\begin{array}{ll} \displaystyle\int \dfrac{4x^2+2x-1}{x^3+x^2} \mathrm{d}x &= -\displaystyle\int x^{-2} \mathrm{d}x + \displaystyle\int \dfrac{1}{x+1} \mathrm{d}x + 3 \displaystyle\int \dfrac{1}{x} \mathrm{d}x \\ &= \dfrac{1}{x} + \log(x+1) + 3\log(x) + C. \end{array}$$