Homework 5

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Mos Kullathon
921425216

1. Evaluate the following integrals

(a) ∫xcos⁡(5x) ⁣dx\displaystyle\int x\cos(5x)\dd x

u=x  ⟹  u′=1v′=cos⁡(5x)  ⟹  v=15sin⁡(5x)∫xcos⁡(5x) ⁣dx=uv−∫vu′ ⁣dx=x5sin⁡(5x)−∫15sin⁡(5x) ⁣dx=x5sin⁡(5x)−15∫sin⁡(5x) ⁣dx=x5sin⁡(5x)−15(−15cos⁡(5x))+c=15sin⁡(5x)+125cos⁡(5x)+c\begin{darray}{cc} u = x &\implies& u' = 1 \\ v' = \cos(5x) &\implies& v = \frac{1}{5}\sin(5x) \end{darray} \\ \begin{align*} \int x\cos(5x)\dd x &= uv - \int vu' \dd x \\ &= \frac{x}{5}\sin(5x) - \int\frac{1}{5}\sin(5x)\dd x \\ &= \frac{x}{5}\sin(5x) - \frac{1}{5}\int\sin(5x)\dd x \\ &= \frac{x}{5}\sin(5x) - \frac{1}{5}\(-\frac{1}{5}\cos(5x)\) + c \\ &= \frac{1}{5}\sin(5x) + \frac{1}{25}\cos(5x) + c \end{align*}

(b) ∫01x2e−x ⁣dx\displaystyle\int_0^1 x^2e^{-x}\dd x

u=x2  ⟹  u′=2xv′=e−x  ⟹  v=−e−x∫01x2e−x ⁣dx=[uv]01−∫01vu′ ⁣dx=[−x2e−x]01−∫01−2xe−x ⁣dx=[−x2e−x]01+2∫01xe−x ⁣dxp=x  ⟹  p′=1q′=e−x  ⟹  q=−e−x∫01xe−x ⁣dx=[pq]01−∫01qp′ ⁣dx=[−xe−x]01−∫01−e−x ⁣dx=[−xe−x]01+[−e−x]01=−1e−1e+1=−2+ee∴∫01x2e−x ⁣dx=[−x2e−x]01+2(−2+ee)=−1e+−4+2ee=−5+2ee\begin{darray}{cc} u = x^2 &\implies& u' = 2x \\ v' = e^{-x} &\implies& v = -e^{-x} \end{darray} \\ \begin{align*} \int_0^1 x^2 e^{-x}\dd x &= \bigg[uv\bigg]_0^1 - \int_0^1 vu' \dd x \\ &= \bigg[-x^2e^{-x}\bigg]_0^1 - \int_0^1 -2xe^{-x} \dd x \\ &= \bigg[-x^2e^{-x}\bigg]_0^1 + 2\int_0^1 xe^{-x} \dd x \end{align*} \\ \begin{darray}{cc} p = x &\implies& p' = 1 \\ q' = e^{-x} &\implies& q = -e^{-x} \end{darray} \\ \begin{align*} \int_0^1 xe^{-x}\dd x &= \bigg[pq\bigg]_0^1 - \int_0^1 qp' \dd x \\ &= \bigg[-xe^{-x}\bigg]_0^1 - \int_0^1 -e^{-x}\dd x \\ &= \bigg[-xe^{-x}\bigg]_0^1 + \bigg[-e^{-x}\bigg]_0^1 \\ &= -\frac{1}{e} - \frac{1}{e} + 1 \\ &= \frac{-2+e}{e} \end{align*} \\ \begin{align*} \therefore\int_0^1 x^2e^{-x}\dd x &= \bigg[-x^2e^{-x}\bigg]_0^1 + 2\(\frac{-2+e}{e}\) \\ &= -\frac{1}{e} + \frac{-4+2e}{e} \\ &= \frac{-5+2e}{e} \end{align*}

(c) ∫x(ln⁡x)2 ⁣dx\displaystyle\int x(\ln x)^2\dd x

u=(ln⁡x)2  ⟹  u′=2xln⁡xv′=x  ⟹  v=12x2∫x(ln⁡x)2 ⁣dx=uv−∫vu′ ⁣dx=12x2ln⁡2x−∫xln⁡x ⁣dxp=ln⁡x  ⟹  p′=1xq′=x  ⟹  q=12x2∫xln⁡x ⁣dx=pq−∫qp′ ⁣dx=12x2ln⁡x−∫ 12x ⁣dx=12x2ln⁡x−14x2+c∴∫x(ln⁡x)2 ⁣dx=12x2ln⁡2x−12x2ln⁡x+14x2+c=14x2(2ln⁡2(x)−2ln⁡(x)+1)+c\begin{darray}{cc} u = (\ln x)^2 &\implies& u' = \frac{2}{x}\ln x \\ v' = x &\implies& v = \frac{1}{2}x^2 \end{darray} \\ \begin{align*} \int x(\ln x)^2 \dd x &= uv - \int vu'\dd x \\ &= \frac{1}{2}x^2\ln^2 x - \int x\ln x \dd x \end{align*} \\ \begin{darray}{cc} p = \ln x &\implies& p' = \frac{1}{x} \\ q' = x &\implies& q = \frac{1}{2}x^2 \end{darray} \\ \begin{align*} \int x \ln x \dd x &= pq - \int qp'\dd x \\ &= \frac{1}{2}x^2\ln x - \int\ \frac{1}{2}x\dd x \\ &= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + c \end{align*} \\ \begin{align*} \therefore \int x(\ln x)^2\dd x &= \frac{1}{2}x^2\ln^2x - \frac{1}{2}x^2\ln x + \frac{1}{4}x^2 + c \\ &= \frac{1}{4}x^2 (2\ln^2(x) - 2\ln(x) + 1) + c \end{align*}

(d) ∫xcsc⁡2x ⁣dx\displaystyle\int x \csc^2 x\dd x

u=x  ⟹  u′=1v′=csc⁡2x  ⟹  v=−cot⁡x∫xcsc⁡2x ⁣dx=uv−∫vu′ ⁣dx=−xcot⁡(x)−∫−cot⁡x ⁣dx=−xcot⁡(x)+ln⁡∣sin⁡x∣+c\begin{darray}{cc} u = x &\implies& u' = 1 \\ v' = \csc^2x &\implies& v = -\cot x \end{darray} \\ \begin{align*} \int x \csc^2x\dd x &= uv - \int vu'\dd x \\ &= -x\cot(x)- \int-\cot x\dd x \\ &= -x\cot(x)+ \ln|\sin x| + c \end{align*}

(e) ∫0π/4excos⁡(2x) ⁣dx\displaystyle\int_0^{\pi/4} e^x\cos(2x)\dd x

u=cos⁡(2x)  ⟹  u′=−2sin⁡(2x)v′=ex  ⟹  v=ex∫excos⁡(2x) ⁣dx=uv−∫vu′ ⁣dx=excos⁡(2x)−∫−2exsin⁡(2x) ⁣dx=excos⁡(2x)+2∫exsin⁡(2x) ⁣dxp=sin⁡(2x)  ⟹  p′=2cos⁡(2x)q′=ex  ⟹  q=ex∫exsin⁡(2x) ⁣dx=pq−∫qp′ ⁣dx=exsin⁡(2x)−∫2excos⁡(2x)=exsin⁡(2x)−2∫excos⁡(2x)∫excos⁡(2x) ⁣dx=excos⁡(2x)+2(exsin⁡(2x)−2∫excos⁡(2x))=excos⁡(2x)+2exsin⁡(2x)−4∫excos⁡(2x)5∫excos⁡(2x) ⁣dx=excos⁡(2x)+2exsin⁡(2x)∫excos⁡(2x) ⁣dx=15(excos⁡(2x)+2exsin⁡(2x))+c∴∫0π/4excos⁡(2x) ⁣dx=[15(excos⁡(2x)+2exsin⁡(2x))]0π/4=15(2eπ/4−1)\begin{darray}{cc} u = \cos(2x) &\implies& u' = -2\sin(2x) \\ v' = e^x &\implies& v = e^x \end{darray} \\ \begin{align*} \int e^x\cos(2x)\dd x &= uv - \int vu'\dd x \\ &= e^x\cos(2x) - \int -2e^x\sin(2x) \dd x \\ &= e^x\cos(2x) + 2\int e^x\sin(2x)\dd x \end{align*} \\ \begin{darray}{cc} p = \sin(2x) &\implies& p' = 2\cos(2x) \\ q' = e^x &\implies& q = e^x \end{darray} \\ \begin{align*} \int e^x\sin(2x)\dd x &= pq - \int qp' \dd x \\ &= e^x\sin(2x) - \int2e^x\cos(2x) \\ &= e^x\sin(2x) -2 \int e^x\cos(2x) \end{align*} \\ \begin{align*} \int e^x\cos(2x)\dd x &= e^x\cos(2x) + 2 \(e^x\sin(2x) -2 \int e^x\cos(2x)\) \\ &= e^x\cos(2x) + 2e^x\sin(2x) - 4\int e^x\cos(2x) \\ 5 \int e^x\cos(2x)\dd x &= e^x\cos(2x) + 2e^x\sin(2x) \\ \int e^x\cos(2x)\dd x &= \frac{1}{5}(e^x\cos(2x) + 2e^x\sin(2x)) + c \end{align*} \\ \begin{align*} \therefore \int_0^{\pi/4} e^x\cos(2x)\dd x &= \[\frac{1}{5}(e^x\cos(2x) + 2e^x\sin(2x))\]_0^{\pi/4} \\ &= \frac{1}{5}(2e^{\pi/4}-1) \end{align*}

(f) ∫0πecos⁡tsin⁡(2t) ⁣dt\displaystyle\int_0^\pi e^{\cos t} \sin(2t)\dd t (Recall sin⁡(2t)=2sin⁡tcos⁡t\sin(2t)=2\sin t\cos t)

u=cos⁡t  ⟹   ⁣du=−sin⁡t ⁣dt  ⟺   ⁣dt=1−sin⁡t ⁣dut=0  ⟹  u=cos⁡0=1t=π  ⟹  u=cos⁡π=−1∫0πecos⁡tsin⁡(2t) ⁣dt=2∫1−1ecos⁡tsin⁡tcos⁡t ⁣dt=2∫1−1ueusin⁡t(1−sin⁡t) ⁣du=−2∫1−1ueu ⁣du=2∫−11ueu ⁣dup=u  ⟹  p′=1q′=eu  ⟹  q=eu∫−11ueu ⁣du=[pq]−11−∫qp′ ⁣du=[ueu]−11−∫eu ⁣du=[ueu]−11−[eu]−11=e+1e−e+1e=2e∴∫0πecos⁡tsin⁡(2t) ⁣dt=2∫1−1ecos⁡tsin⁡tcos⁡t ⁣dt=2⋅2e=4e\begin{darray}{cc} u = \cos t &\implies& \dd u = -\sin t\dd t \\ &\iff& \dd t = \frac{1}{-\sin t}\dd u \\ t = 0 &\implies& u = \cos 0 = 1 \\ t = \pi &\implies& u = \cos\pi = -1 \end{darray} \\ \begin{align*} \int_0^\pi e^{\cos t}\sin(2t)\dd t &= 2 \int_1^{-1} e^{\cos t}\sin t\cos t \dd t \\ &= 2\int_1^{-1} ue^u\cancel{\sin t} \(\frac{1}{\cancel{-\sin t}}\)\dd u \\ &= -2\int_1^{-1} ue^u\dd u \\ &= 2\int_{-1}^1 ue^u\dd u \end{align*} \\ \begin{darray}{cc} p = u &\implies& p' = 1 \\ q' = e^u &\implies& q = e^u \end{darray} \\ \begin{align*} \int_{-1}^1 ue^u\dd u &= \bigg[pq\bigg]_{-1}^1 - \int qp' \dd u \\ &= \bigg[ue^u\bigg]_{-1}^1 - \int e^u \dd u \\ &= \bigg[ue^u\bigg]_{-1}^1 - \bigg[e^u\bigg]_{-1}^1 \\ &= e + \frac{1}{e} - e + \frac{1}{e} \\ &= \frac{2}{e} \\ \end{align*} \\ \begin{align*} \therefore \int_0^\pi e^{\cos t}\sin(2t)\dd t &= 2 \int_1^{-1} e^{\cos t}\sin t\cos t \dd t \\ &= 2 \cdot\frac{2}{e} \\ &= \frac{4}{e} \end{align*}

2. Show that the reduction formula for In=∫(ln⁡x)n ⁣dx\displaystyle I_n = \int(\ln x)^n \dd x is In=x(ln⁡x)n−nIn−1I_n = x (\ln x)^n - nI_{n-1}

u=(ln⁡x)n  ⟹  u′=nx(ln⁡x)n−1v′=1  ⟹  v=xIn=∫(ln⁡x)n=uv−∫vu′ ⁣dx=x(ln⁡x)n−∫nxx(ln⁡x)n−1=x(ln⁡x)n−n∫(ln⁡x)n−1⏟In−1=x(ln⁡x)n−nIn−1\begin{darray}{cc} u = (\ln x)^n &\implies& u' = \frac{n}{x}(\ln x)^{n-1} \\ v' = 1 &\implies& v = x \end{darray} \\ \begin{align*} I_n = \int (\ln x)^n &= uv - \int vu'\dd x \\ &= x(\ln x)^n - \int\frac{n\cancel{x}}{\cancel{x}}(\ln x)^{n-1} \\ &= x(\ln x)^n - n\underbrace{\int (\ln x)^{n-1}}_{I_{n-1}} \\ &= x(\ln x)^n - nI_{n-1} \end{align*}

Book Section 3.3:

(135) ∫ ⁣dxx2−a2 ⁣dx\displaystyle\int\frac{\dd x}{\sqrt{x^2-a^2}}\dd x

∫ ⁣dxx2−a2 ⁣dx=∫1x2−a2 ⁣dxx=asec⁡θ  ⟹   ⁣dx=asec⁡θtan⁡θ ⁣dθ  ⟺  θ=sec⁡−1(xa)∫1x2−a2 ⁣dx=∫1a2sec⁡2θ−a2(asec⁡θtan⁡θ ⁣dθ)=∫1a2(sec⁡2θ−1)(asec⁡θtan⁡θ ⁣dθ)=∫1a2tan⁡2θ(asec⁡θtan⁡θ ⁣dθ)=∫1atan⁡θ(asec⁡θtan⁡θ ⁣dθ)=∫sec⁡θ ⁣dθ=ln⁡∣tan⁡θ+sec⁡θ∣+c=ln⁡∣tan⁡(sec⁡−1(xa))+sec⁡(sec⁡−1(xa))∣+c=ln⁡∣tan⁡(sec⁡−1(xa))+xa∣+c=ln⁡∣xa1−1x2a2+xa∣+c=ln⁡∣xax2x2−a2x2+xa∣+c=ln⁡∣xa(1x2)x2−a2+xa∣+c=ln⁡∣1ax2−a2+xa∣+c=ln⁡∣1a(x2−a2+x)∣+c=ln⁡∣1a∣+ln⁡∣x2−a2+x∣+c=ln⁡∣x2−a2+x∣+c\int\frac{\dd x}{\sqrt{x^2-a^2}}\dd x = \int\frac{1}{\sqrt{x^2-a^2}}\dd x \\ \begin{darray}{cc} x = a\sec\theta &\implies& \dd x = a\sec\theta\tan\theta\dd\theta \\ &\iff& \theta = \sec^{-1}\(\frac{x}{a}\) \end{darray} \\ \begin{align*} \int\frac{1}{\sqrt{x^2-a^2}}\dd x &= \int\frac{1}{\sqrt{a^2\sec^2\theta - a^2}} (a\sec\theta\tan\theta\dd\theta) \\ &= \int\frac{1}{\sqrt{a^2(\sec^2\theta - 1)}} (a\sec\theta\tan\theta\dd\theta) \\ &= \int\frac{1}{\sqrt{a^2\tan^2\theta}} (a\sec\theta\tan\theta\dd\theta) \\ &= \int\frac{1}{\cancel{a\tan\theta}} (\cancel{a}\sec\theta\cancel{\tan\theta}\dd\theta) \\ &= \int\sec\theta\dd\theta \\ &= \ln|\tan\theta + \sec\theta| + c \\ &= \ln\left|\tan\(\sec^{-1}\(\frac{x}{a}\)\) + \sec\(\sec^{-1}\(\frac{x}{a}\)\)\right| + c \\ &= \ln\left|\tan\(\sec^{-1}\(\frac{x}{a}\)\) + \frac{x}{a}\right| + c \\ &= \ln|\frac{x}{a}\sqrt{1-\frac{1}{\frac{x^2}{a^2}}} + \frac{x}{a}| + c \\ &= \ln\left|\frac{x}{a}\sqrt{\frac{x^2}{x^2} - \frac{a^2}{x^2}} + \frac{x}{a}\right| + c \\ &= \ln\left|\frac{x}{a}\sqrt{\(\frac{1}{x^2}\)x^2-a^2} + \frac{x}{a}\right| + c \\ &= \ln\left|\frac{1}{a}\sqrt{x^2-a^2} + \frac{x}{a}\right| + c \\ &= \ln\left|\frac{1}{a}(\sqrt{x^2-a^2} + x) \right| + c \\ &= \ln\left|\frac{1}{a}\right| + \ln|\sqrt{x^2-a^2} + x| + c \\ &= \ln|\sqrt{x^2-a^2} + x| + c \end{align*}

(140) ∫ ⁣dx(1+x2)2\displaystyle\int\frac{\dd x}{(1+x^2)^2}

∫ ⁣dx(1+x2)2=∫1(1+x2)2 ⁣dxx=tan⁡θ  ⟹   ⁣dx=sec⁡2θ ⁣dθ  ⟺  θ=tan⁡−1x  ⟺  sin⁡θ=x1+x2  ⟺  cos⁡θ=11+x2∫1(1+x2)2 ⁣dx=∫1(1+tan⁡2θ)2(sec⁡2θ ⁣dθ)=∫1sec⁡4θ(sec⁡2 ⁣dθ)=∫1sec⁡2θ ⁣dθ=∫cos⁡2θ ⁣dθ=∫12(cos⁡(2θ)+1) ⁣dθ=12∫cos⁡(2θ)+1 ⁣dθ=12(θ+12sin⁡2θ)+c=12(θ+122sin⁡θcos⁡θ)+c=12(θ+sin⁡θcos⁡θ)+c=12(tan⁡−1(x)+x1+x2)+c\int\frac{\dd x}{(1+x^2)^2} = \int\frac{1}{(1+x^2)^2}\dd x \\ \begin{darray}{cc} x = \tan\theta &\implies& \dd x = \sec^2\theta\dd\theta \\ &\iff& \theta = \tan^{-1}x \\ &\iff& \sin\theta = \frac{x}{\sqrt{1+x^2}} \\ &\iff& \cos\theta = \frac{1}{\sqrt{1+x^2}} \end{darray} \\ \begin{align*} \int\frac{1}{(1+x^2)^2}\dd x &= \int\frac{1}{(1 + \tan^2\theta)^2} (\sec^2\theta\dd\theta) \\ &= \int\frac{1}{\sec^4\theta}(\sec^2\dd\theta) \\ &= \int\frac{1}{\sec^2\theta}\dd\theta \\ &= \int\cos^2\theta\dd\theta \\ &= \int\frac{1}{2}(\cos(2\theta) + 1)\dd\theta \\ &= \frac{1}{2}\int\cos(2\theta) + 1 \dd\theta \\ &= \frac{1}{2}\(\theta + \frac{1}{2}\sin2\theta\) + c \\ &= \frac{1}{2}\(\theta+\frac{1}{2}2\sin\theta\cos\theta\) + c \\ &= \frac{1}{2}(\theta + \sin\theta\cos\theta) + c \\ &= \frac{1}{2} \(\tan^{-1}(x) + \frac{x}{1+x^2}\) + c \end{align*}

(142) ∫x2−25x ⁣dx\displaystyle\int\frac{\sqrt{x^2-25}}{x}\dd x

r2=25  ⟹  r=5x=rsec⁡θ  ⟹   ⁣dx=rsec⁡θtan⁡θ ⁣dθ  ⟺  θ=sec⁡−1(xr)  ⟺  sec⁡θ=xr  ⟺  tan⁡θ=x2−r2r∫x2−25x ⁣dx=∫r2sec⁡2θ−r2rsec⁡θ(rsec⁡θtan⁡θ ⁣dθ)=∫r2(sec⁡2θ−1)rsec⁡θ(rsec⁡θtan⁡θ ⁣dθ)=∫r2tan⁡2θrsec⁡θ(rsec⁡θtan⁡θ ⁣dθ)=∫rtan⁡θrsec⁡θ(rsec⁡θtan⁡θ ⁣dθ)=r∫tan⁡2θ ⁣dθ=r∫sin⁡2θcos⁡2θ ⁣dθ=r∫1−cos⁡2θcos⁡2θ ⁣dθ=r∫1cos⁡2θ−1 ⁣dθ=r∫sec⁡2(θ)−1 ⁣dθ=r(tan⁡(θ)−θ)+c=r(x2−r2r−sec⁡−1(xr))+c=x2−r2−rsec⁡−1(xr)+c=x2−25−5sec⁡−1(x5)+c\begin{darray}{cc} r^2 = 25 &\implies& r = 5 \\ x = r\sec\theta &\implies& \dd x = r\sec\theta\tan\theta\dd\theta \\ &\iff& \theta = \sec^{-1}\(\frac{x}{r}\) \\ &\iff& \sec\theta = \frac{x}{r} \\ &\iff& \tan\theta = \frac{\sqrt{x^2 - r^2}}{r} \end{darray} \\ \begin{align*} \int\frac{\sqrt{x^2-25}}{x}\dd x &= \int\frac{\sqrt{r^2\sec^2\theta - r^2}}{r\sec\theta} (r\sec\theta\tan\theta\dd\theta) \\ &= \int\frac{\sqrt{r^2(\sec^2\theta - 1)}}{r\sec\theta} (r\sec\theta\tan\theta\dd\theta) \\ &= \int\frac{\sqrt{r^2\tan^2\theta}}{r\sec\theta} (r\sec\theta\tan\theta\dd\theta) \\ &= \int\frac{r\tan\theta}{\cancel{r\sec}\theta} (\cancel{r\sec}\theta\tan\theta\dd\theta) \\ &= r\int\tan^2\theta\dd\theta \\ &= r\int\frac{\sin^2\theta}{\cos^2\theta}\dd\theta \\ &= r\int\frac{1-\cos^2\theta}{\cos^2\theta}\dd\theta \\ &= r\int\frac{1}{\cos^2\theta} - 1 \dd\theta \\ &= r\int\sec^2(\theta) - 1 \dd\theta \\ &= r(\tan(\theta)-\theta) + c \\ &= r\(\frac{\sqrt{x^2 - r^2}}{r} - \sec^{-1}\(\frac{x}{r}\)\) + c \\ &= \sqrt{x^2 - r^2} - r\sec^{-1}\(\frac{x}{r}\) + c \\ &= \sqrt{x^2 - 25} - 5\sec^{-1}\(\frac{x}{5}\) + c \end{align*}

(153) ∫−11(1−x2)3/2 ⁣dx\displaystyle\int_{-1}^1 (1-x^2)^{3/2}\dd x

∫−11(1−x2)3/2 ⁣dx=∫−11(1−x2)3 ⁣dxx=sin⁡θ  ⟹   ⁣dx=cos⁡θ ⁣dθ  ⟺  θ=sin⁡−1(x)x=sin⁡θ=−1  ⟹  θ=−π2x=sin⁡θ=1  ⟹  θ=π2∫−11(1−x2)3 ⁣dx=∫−π/2π/2(1−sin⁡2θ)3⋅(cos⁡θ ⁣dθ)=∫−π/2π/2(cos⁡2θ)3⋅(cos⁡θ ⁣dθ)=∫−π/2π/2(cos⁡2θ)2cos⁡2θ⋅(cos⁡θ ⁣dθ)=∫−π/2π/2cos⁡2θcos⁡θ⋅(cos⁡θ ⁣dθ)=∫−π/2π/2cos⁡2θcos⁡2θ ⁣dθ=∫−π/2π/212(cos⁡(2θ)+1)⋅12(cos⁡(2θ)+1) ⁣dθ=14∫−π/2π/2(cos⁡(2θ)+1)2 ⁣dθ=14∫−π/2π/2cos⁡2(2θ)+1+2cos⁡(2θ) ⁣dθ=14∫−π/2π/212(cos⁡(4θ)+1)+1+2cos⁡(2θ) ⁣dθ=14∫−π/2π/212cos⁡(4θ)+12+1+2cos⁡(2θ) ⁣dθ=14∫−π/2π/212(cos⁡(4θ)+1+2+4cos⁡(2θ)) ⁣dθ=18∫−π/2π/2(cos⁡(4θ)+4cos⁡(2θ)+3) ⁣dθ=18[sin⁡(4θ)4+4sin⁡(2θ)+3θ]−π/2π/2=18(32π−(−32π))=38π\int_{-1}^1 (1-x^2)^{3/2}\dd x = \int_{-1}^1 \sqrt{(1-x^2)^3}\dd x \\ \begin{darray}{cc} x = \sin\theta &\implies& \dd x= \cos\theta\dd\theta \\ &\iff& \theta = \sin^{-1}(x) \\ x = \sin\theta = -1 &\implies& \theta = -\frac{\pi}{2} \\ x = \sin\theta = 1 &\implies& \theta = \frac{\pi}{2} \end{darray} \\ \begin{align*} \int_{-1}^1\sqrt{(1-x^2)^3}\dd x &= \int_{-\pi/2}^{\pi/2}\sqrt{(1-\sin^2\theta)^3} \cdot(\cos\theta\dd\theta) \\ &= \int_{-\pi/2}^{\pi/2}\sqrt{(\cos^2\theta)^3} \cdot(\cos\theta\dd\theta) \\ &= \int_{-\pi/2}^{\pi/2}\sqrt{(\cos^2\theta)^2\cos^2\theta} \cdot(\cos\theta\dd\theta) \\ &= \int_{-\pi/2}^{\pi/2}\cos^2\theta\cos\theta \cdot(\cos\theta\dd\theta) \\ &= \int_{-\pi/2}^{\pi/2}\cos^2\theta\cos^2\theta\dd\theta \\ &= \int_{-\pi/2}^{\pi/2}\frac{1}{2}(\cos(2\theta)+1)\cdot\frac{1}{2}(\cos(2\theta)+1)\dd\theta \\ &= \frac{1}{4}\int_{-\pi/2}^{\pi/2}(\cos(2\theta)+1)^2\dd\theta \\ &= \frac{1}{4}\int_{-\pi/2}^{\pi/2}\cos^2(2\theta) + 1 + 2\cos(2\theta)\dd\theta \\ &= \frac{1}{4}\int_{-\pi/2}^{\pi/2}\frac{1}{2}(\cos(4\theta) + 1) + 1 + 2\cos(2\theta)\dd\theta \\ &= \frac{1}{4}\int_{-\pi/2}^{\pi/2}\frac{1}{2}\cos(4\theta) + \frac{1}{2} + 1 + 2\cos(2\theta)\dd\theta \\ &= \frac{1}{4}\int_{-\pi/2}^{\pi/2}\frac{1}{2}(\cos(4\theta) + 1 + 2 + 4\cos(2\theta))\dd\theta \\ &= \frac{1}{8}\int_{-\pi/2}^{\pi/2}(\cos(4\theta)+4\cos(2\theta)+3)\dd\theta \\ &= \frac{1}{8}\[\frac{\sin(4\theta)}{4}+4\sin(2\theta)+3\theta\]_{-\pi/2}^{\pi/2} \\ &= \frac{1}{8}\(\frac{3}{2}\pi - \(-\frac{3}{2}\pi\)\) \\ &= \frac{3}{8}\pi \end{align*}

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