Exercise 7.8 Practice

Step 1: Find Indefinite Integral
Let $I = \int_{-1}^1 (x+1) dx$.
$\int (x+1) dx = \frac{x^2}{2} + x = F(x)$

Step 2: Apply Limits $F(b) - F(a)$
$F(1) = \frac{1^2}{2} + 1 = \frac{3}{2}$
$F(-1) = \frac{(-1)^2}{2} + (-1) = \frac{1}{2} - 1 = -\frac{1}{2}$

Step 3: Final Calculation
$I = F(1) - F(-1) = \frac{3}{2} - (-\frac{1}{2}) = \frac{4}{2} = 2$

$\int (x^2+3) dx = \frac{x^3}{3} + 3x$.

Upper limit (2): $\frac{8}{3} + 6 = \frac{26}{3}$.

Lower limit (0): $0$.

Step 1: Find Indefinite Integral
$\int \frac{1}{x} dx = \log|x| = F(x)$

Step 2: Apply Limits
$I = F(3) - F(2) = \log 3 - \log 2$

Step 3: Simplify
Using $\log a - \log b = \log(\frac{a}{b})$:
$I = \log(\frac{3}{2})$

$\int \sec^2 x dx = \tan x$.

Upper limit ($\pi/4$): $\tan(\pi/4) = 1$.

Lower limit (0): $\tan(0) = 0$.

Step 1: Find Indefinite Integral
$\int (4x^3 - 5x^2 + 6x + 9) dx = x^4 - \frac{5x^3}{3} + 3x^2 + 9x = F(x)$

Step 2: Calculate F(2) and F(1)
$F(2) = 2^4 - \frac{5(8)}{3} + 3(4) + 9(2) = 16 - \frac{40}{3} + 12 + 18 = 46 - \frac{40}{3} = \frac{138-40}{3} = \frac{98}{3}$
$F(1) = 1 - \frac{5}{3} + 3 + 9 = 13 - \frac{5}{3} = \frac{39-5}{3} = \frac{34}{3}$

Step 3: Final Answer
$I = F(2) - F(1) = \frac{98}{3} - \frac{34}{3} = \frac{64}{3}$

$\int \frac{1}{x} dx = \log|x|$.

Upper limit ($e$): $\log e = 1$.

Lower limit (1): $\log 1 = 0$.

Step 1: Find Indefinite Integral
$\int \sin 2x dx = -\frac{\cos 2x}{2} = F(x)$

Step 2: Apply Limits
$I = F(\frac{\pi}{4}) - F(0) = -\frac{1}{2}[\cos(2 \cdot \frac{\pi}{4}) - \cos(0)]$

Step 3: Calculate
$= -\frac{1}{2}[\cos(\frac{\pi}{2}) - 1] = -\frac{1}{2}[0 - 1] = \frac{1}{2}$

$\int e^{2x} dx = \frac{e^{2x}}{2}$.

Upper limit (1): $\frac{e^2}{2}$.

Lower limit (0): $\frac{e^0}{2} = \frac{1}{2}$.

Step 1: Find Indefinite Integral
$\int \cos 2x dx = \frac{\sin 2x}{2} = F(x)$

Step 2: Apply Limits
$I = [\frac{\sin 2x}{2}]_0^{\pi/2} = \frac{1}{2}(\sin \pi - \sin 0)$

Step 3: Calculate
$= \frac{1}{2}(0 - 0) = 0$

Let $1+x^2=t \Rightarrow 2x dx = dt$.

$\int \frac{1}{t} dt = \log|t|$.

Limits: $x=0 \to t=1$, $x=1 \to t=2$.

$\log 2 - \log 1 = \log 2$.

Step 1: Find Indefinite Integral
$\int e^x dx = e^x = F(x)$

Step 2: Apply Limits
$I = F(5) - F(4) = e^5 - e^4$

Step 3: Final Answer
$= e^4(e - 1)$

Step 1: Find Indefinite Integral
$\int \tan x dx = -\log|\cos x| = F(x)$

Step 2: Apply Limits
$I = [-\log|\cos x|]_0^{\pi/4} = -\log(\cos \frac{\pi}{4}) - (-\log(\cos 0))$

Step 3: Calculate
$= -\log(\frac{1}{\sqrt{2}}) + \log(1) = -(\log 1 - \log \sqrt{2}) + 0$
$= \log \sqrt{2} = \frac{1}{2}\log 2$

Step 1: Find Indefinite Integral
$\int \csc x dx = \log|\csc x - \cot x| = F(x)$

Step 2: Apply Limits
$F(\frac{\pi}{4}) = \log|\csc \frac{\pi}{4} - \cot \frac{\pi}{4}| = \log|\sqrt{2} - 1|$
$F(\frac{\pi}{6}) = \log|\csc \frac{\pi}{6} - \cot \frac{\pi}{6}| = \log|2 - \sqrt{3}|$

Step 3: Final Answer
$I = \log(\sqrt{2}-1) - \log(2-\sqrt{3}) = \log\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)$

Step 1: Find Indefinite Integral
$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x = F(x)$

Step 2: Apply Limits
$I = \sin^{-1}(1) - \sin^{-1}(0)$

Step 3: Calculate
$= \frac{\pi}{2} - 0 = \frac{\pi}{2}$

Step 1: Find Indefinite Integral
$\int \frac{dx}{1+x^2} = \tan^{-1} x = F(x)$

Step 2: Apply Limits
$I = \tan^{-1}(1) - \tan^{-1}(0)$

Step 3: Calculate
$= \frac{\pi}{4} - 0 = \frac{\pi}{4}$

Step 1: Find Indefinite Integral
Using $\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log|\frac{x-a}{x+a}|$:
$\int \frac{dx}{x^2-1} = \frac{1}{2}\log|\frac{x-1}{x+1}| = F(x)$

Step 2: Apply Limits
$F(3) = \frac{1}{2}\log|\frac{2}{4}| = \frac{1}{2}\log(\frac{1}{2})$
$F(2) = \frac{1}{2}\log|\frac{1}{3}| = \frac{1}{2}\log(\frac{1}{3})$

Step 3: Calculate
$I = \frac{1}{2}[\log(\frac{1}{2}) - \log(\frac{1}{3})] = \frac{1}{2}\log(\frac{1/2}{1/3}) = \frac{1}{2}\log(\frac{3}{2})$

Step 1: Use Identity
$\cos^2 x = \frac{1+\cos 2x}{2}$

Step 2: Integrate
$\int (\frac{1}{2} + \frac{1}{2}\cos 2x) dx = \frac{x}{2} + \frac{1}{4}\sin 2x = F(x)$

Step 3: Apply Limits
$F(\frac{\pi}{2}) = \frac{\pi}{4} + \frac{1}{4}\sin \pi = \frac{\pi}{4} + 0$
$F(0) = 0 + 0 = 0$
$I = \frac{\pi}{4}$

Step 1: Substitution
Let $x^2+1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}$.
Indefinite integral: $\frac{1}{2}\log|x^2+1| = F(x)$

Step 2: Apply Limits
$F(3) = \frac{1}{2}\log(10)$
$F(2) = \frac{1}{2}\log(5)$

Step 3: Calculate
$I = \frac{1}{2}(\log 10 - \log 5) = \frac{1}{2}\log(\frac{10}{5}) = \frac{1}{2}\log 2$

Step 1: Split Integral
$I = \int_0^1 \frac{2x}{5x^2+1} dx + 3\int_0^1 \frac{1}{5x^2+1} dx = I_1 + I_2$

Step 2: Solve $I_1$
Let $5x^2+1=t \Rightarrow 10x dx = dt \Rightarrow 2x dx = \frac{dt}{5}$.
$I_1 = \frac{1}{5}[\log|5x^2+1|]_0^1 = \frac{1}{5}(\log 6 - \log 1) = \frac{1}{5}\log 6$

Step 3: Solve $I_2$
$I_2 = 3\int_0^1 \frac{dx}{(\sqrt{5}x)^2+1} = 3[\frac{1}{\sqrt{5}}\tan^{-1}(\sqrt{5}x)]_0^1$
$= \frac{3}{\sqrt{5}}(\tan^{-1}\sqrt{5} - 0) = \frac{3}{\sqrt{5}}\tan^{-1}\sqrt{5}$

Step 1: Substitution
Let $x^2 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}$.
Limits: When $x=0, t=0$. When $x=1, t=1$.

Step 2: Integrate
$I = \frac{1}{2} \int_0^1 e^t dt = \frac{1}{2} [e^t]_0^1$

Step 3: Calculate
$= \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1)$

Step 1: Simplify Integrand
Divide $5x^2$ by $x^2+4x+3$: Quotient 5, Remainder $-(20x+15)$.
$\frac{5x^2}{x^2+4x+3} = 5 - \frac{20x+15}{(x+1)(x+3)}$

Step 2: Partial Fractions
$\frac{20x+15}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}$
$20x+15 = A(x+3) + B(x+1)$.
Put $x=-1 \Rightarrow -5 = 2A \Rightarrow A = -5/2$.
Put $x=-3 \Rightarrow -45 = -2B \Rightarrow B = 45/2$.

Step 3: Integrate
$I = \int_1^2 [5 - (-\frac{5}{2(x+1)} + \frac{45}{2(x+3)})] dx$
$= [5x + \frac{5}{2}\log(x+1) - \frac{45}{2}\log(x+3)]_1^2$

Step 4: Apply Limits
Upper: $10 + \frac{5}{2}\log 3 - \frac{45}{2}\log 5$
Lower: $5 + \frac{5}{2}\log 2 - \frac{45}{2}\log 4$
$I = 5 + \frac{5}{2}(\log 3 - \log 2) - \frac{45}{2}(\log 5 - \log 4)$

Step 1: Find Indefinite Integral
$\int (2\sec^2 x + x^3 + 2) dx = 2\tan x + \frac{x^4}{4} + 2x = F(x)$

Step 2: Apply Limits
$F(\frac{\pi}{4}) = 2\tan(\frac{\pi}{4}) + \frac{(\pi/4)^4}{4} + 2(\frac{\pi}{4})$
$= 2(1) + \frac{\pi^4}{1024} + \frac{\pi}{2}$
$F(0) = 0$

Step 1: Use Identity
$\cos^2 A - \sin^2 A = \cos 2A$.
So, $\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos x$.

Step 2: Integrate
$I = \int_0^\pi -\cos x dx = -[\sin x]_0^\pi$

Step 3: Calculate
$= -(\sin \pi - \sin 0) = -(0 - 0) = 0$

Step 1: Split Integral
$I = \int_0^2 \frac{6x}{x^2+4} dx + \int_0^2 \frac{3}{x^2+4} dx$
$= 3\int_0^2 \frac{2x}{x^2+4} dx + 3\int_0^2 \frac{1}{x^2+2^2} dx$

Step 2: Integrate
$= 3[\log(x^2+4)]_0^2 + 3[\frac{1}{2}\tan^{-1}\frac{x}{2}]_0^2$

Step 3: Apply Limits
$= 3(\log 8 - \log 4) + \frac{3}{2}(\tan^{-1} 1 - \tan^{-1} 0)$
$= 3\log 2 + \frac{3}{2}(\frac{\pi}{4})$

Step 1: Integrate First Term (Parts)
$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x$.
Limits $0 \to 1$: $(e-e) - (0-1) = 1$.

Step 2: Integrate Second Term
$\int_0^1 \sin \frac{\pi x}{4} dx = [-\frac{4}{\pi}\cos \frac{\pi x}{4}]_0^1$
$= -\frac{4}{\pi}(\cos \frac{\pi}{4} - \cos 0) = -\frac{4}{\pi}(\frac{1}{\sqrt{2}} - 1)$

Step 3: Combine
$I = 1 - \frac{2\sqrt{2}}{\pi} + \frac{4}{\pi}$

Step 1: Integrate
$\int \frac{dx}{1+x^2} = \tan^{-1} x$.

Step 2: Apply Limits
$[\tan^{-1} x]_1^{\sqrt{3}} = \tan^{-1}\sqrt{3} - \tan^{-1} 1$
$= \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$

Step 1: Simplify Integral
$\int \frac{dx}{4+9x^2} = \frac{1}{9} \int \frac{dx}{(2/3)^2 + x^2}$

Step 2: Integrate
Using $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$ with $a=2/3$:
$= \frac{1}{9} [\frac{1}{2/3} \tan^{-1}(\frac{x}{2/3})] = \frac{1}{6} \tan^{-1}(\frac{3x}{2})$

Step 3: Apply Limits
$\frac{1}{6} [\tan^{-1}(\frac{3(2/3)}{2}) - \tan^{-1} 0] = \frac{1}{6} [\tan^{-1} 1 - 0]$
$= \frac{1}{6} (\frac{\pi}{4}) = \frac{\pi}{24}$