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SESSCALCET2 6.1.011.MI.
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Evaluate the integral. (Use $C$ for the constant of integration.)
$\int \arctan 6 t d t$
$\square$
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SESSCALCET2 6.1.013.MI.
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Evaluate the integral. (Use $C$ for the constant of integration.)
$$
\int e^{2 \theta} \sin (3 \theta) d \theta
$$
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SESSCALCET2 6.1.021.
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Evaluate the integral.
$$
\int_{0}^{1} 8 t \cosh t d t
$$
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∻Solution by Steps∻ ### Question 1: Evaluate the integral $\int \arctan 6t \, dt$ ‖ step 1 ⋮ Let $I = \int \arctan 6t \, dt$. ‖ ‖ step 2 ⋮ Use integration by parts: Let $u = \arctan 6t$ and $dv = dt$. ‖ ‖ step 3 ⋮ Compute $du$ and $v$: $du = \frac{6}{1 + (6t)^2} \, dt$ and $v = t$. ‖ ‖ step 4 ⋮ Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. ‖ ‖ step 5 ⋮ Substitute $u$, $v$, $du$, and $dv$ into the formula: $\int \arctan 6t \, dt = t \arctan 6t - \int t \cdot \frac{6}{1 + 36t^2} \, dt$. ‖ ‖ step 6 ⋮ Simplify the integral: $\int t \cdot \frac{6}{1 + 36t^2} \, dt = \int \frac{6t}{1 + 36t^2} \, dt$. ‖ ‖ step 7 ⋮ Use substitution: Let $u = 36t^2 + 1$, then $du = 72t \, dt$. ‖ ‖ step 8 ⋮ Rewrite the integral: $\int \frac{6t}{1 + 36t^2} \, dt = \int \frac{6t}{u} \cdot \frac{du}{72t} = \frac{1}{12} \int \frac{1}{u} \, du$. ‖ ‖ step 9 ⋮ Integrate: $\frac{1}{12} \int \frac{1}{u} \, du = \frac{1}{12} \ln |u| + C = \frac{1}{12} \ln |36t^2 + 1| + C$. ‖ ‖ step 10 ⋮ Combine results: $\int \arctan 6t \, dt = t \arctan 6t - \frac{1}{12} \ln |36t^2 + 1| + C$. ‖ ∻Answer∻ ⚹ $\int \arctan 6t \, dt = t \arctan 6t - \frac{1}{12} \ln |36t^2 + 1| + C$ ⚹ ∻Key Concept∻ ⚹Integration by parts⚹ ∻Explanation∻ ⚹Integration by parts is used to integrate the product of two functions.⚹ ### Question 2: Evaluate the integral $\int e^{2 heta} \sin(3 heta) \, d heta$ ‖ step 1 ⋮ Let $I = \int e^{2 heta} \sin(3 heta) \, d heta$. ‖ ‖ step 2 ⋮ Use integration by parts twice: Let $u = e^{2 heta}$ and $dv = \sin(3 heta) \, d heta$. ‖ ‖ step 3 ⋮ Compute $du$ and $v$: $du = 2e^{2 heta} \, d heta$ and $v = -\frac{1}{3} \cos(3 heta)$. ‖ ‖ step 4 ⋮ Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. ‖ ‖ step 5 ⋮ Substitute $u$, $v$, $du$, and $dv$ into the formula: $\int e^{2 heta} \sin(3 heta) \, d heta = -\frac{1}{3} e^{2 heta} \cos(3 heta) - \int -\frac{1}{3} \cos(3 heta) \cdot 2e^{2 heta} \, d heta$. ‖ ‖ step 6 ⋮ Simplify the integral: $\int e^{2 heta} \sin(3 heta) \, d heta = -\frac{1}{3} e^{2 heta} \cos(3 heta) + \frac{2}{3} \int e^{2 heta} \cos(3 heta) \, d heta$. ‖ ‖ step 7 ⋮ Use integration by parts again: Let $u = e^{2 heta}$ and $dv = \cos(3 heta) \, d heta$. ‖ ‖ step 8 ⋮ Compute $du$ and $v$: $du = 2e^{2 heta} \, d heta$ and $v = \frac{1}{3} \sin(3 heta)$. ‖ ‖ step 9 ⋮ Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. ‖ ‖ step 10 ⋮ Substitute $u$, $v$, $du$, and $dv$ into the formula: $\int e^{2 heta} \cos(3 heta) \, d heta = \frac{1}{3} e^{2 heta} \sin(3 heta) - \int \frac{1}{3} \sin(3 heta) \cdot 2e^{2 heta} \, d heta$. ‖ ‖ step 11 ⋮ Simplify the integral: $\int e^{2 heta} \cos(3 heta) \, d heta = \frac{1}{3} e^{2 heta} \sin(3 heta) - \frac{2}{3} \int e^{2 heta} \sin(3 heta) \, d heta$. ‖ ‖ step 12 ⋮ Combine results: $\int e^{2 heta} \sin(3 heta) \, d heta = -\frac{1}{3} e^{2 heta} \cos(3 heta) + \frac{2}{3} \left( \frac{1}{3} e^{2 heta} \sin(3 heta) - \frac{2}{3} \int e^{2 heta} \sin(3 heta) \, d heta ight)$. ‖ ‖ step 13 ⋮ Solve for the integral: $I = -\frac{1}{3} e^{2 heta} \cos(3 heta) + \frac{2}{9} e^{2 heta} \sin(3 heta) - \frac{4}{9} I$. ‖ ‖ step 14 ⋮ Combine like terms: $I + \frac{4}{9} I = -\frac{1}{3} e^{2 heta} \cos(3 heta) + \frac{2}{9} e^{2 heta} \sin(3 heta)$. ‖ ‖ step 15 ⋮ Simplify: $\frac{13}{9} I = -\frac{1}{3} e^{2 heta} \cos(3 heta) + \frac{2}{9} e^{2 heta} \sin(3 heta)$. ‖ ‖ step 16 ⋮ Solve for $I$: $I = \frac{9}{13} \left( -\frac{1}{3} e^{2 heta} \cos(3 heta) + \frac{2}{9} e^{2 heta} \sin(3 heta) ight) + C$. ‖ ‖ step 17 ⋮ Simplify: $I = -\frac{3}{13} e^{2 heta} \cos(3 heta) + \frac{2}{13} e^{2 heta} \sin(3 heta) + C$. ‖ ∻Answer∻ ⚹ $\int e^{2 heta} \sin(3 heta) \, d heta = -\frac{3}{13} e^{2 heta} \cos(3 heta) + \frac{2}{13} e^{2 heta} \sin(3 heta) + C$ ⚹ ∻Key Concept∻ ⚹Integration by parts⚹ ∻Explanation∻ ⚹Integration by parts is used to integrate the product of two functions, and sometimes it needs to be applied more than once.⚹ ### Question 3: Evaluate the integral $\int_{0}^{1} 8t \cosh t \, dt$ ‖ step 1 ⋮ Let $I = \int_{0}^{1} 8t \cosh t \, dt$. ‖ ‖ step 2 ⋮ Use integration by parts: Let $u = 8t$ and $dv = \cosh t \, dt$. ‖ ‖ step 3 ⋮ Compute $du$ and $v$: $du = 8 \, dt$ and $v = \sinh t$. ‖ ‖ step 4 ⋮ Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. ‖ ‖ step 5 ⋮ Substitute $u$, $v$, $du$, and $dv$ into the formula: $\int_{0}^{1} 8t \cosh t \, dt = \left[ 8t \sinh t ight]_{0}^{1} - \int_{0}^{1} 8 \sinh t \, dt$. ‖ ‖ step 6 ⋮ Evaluate the boundary term: $\left[ 8t \sinh t ight]_{0}^{1} = 8 \sinh 1 - 0 = 8 \sinh 1$. ‖ ‖ step 7 ⋮ Simplify the integral: $\int_{0}^{1} 8 \sinh t \, dt = 8 \int_{0}^{1} \sinh t \, dt$. ‖ ‖ step 8 ⋮ Integrate: $\int \sinh t \, dt = \cosh t$. ‖ ‖ step 9 ⋮ Evaluate the boundary term: $8 \left[ \cosh t ight]_{0}^{1} = 8 (\cosh 1 - \cosh 0) = 8 (\cosh 1 - 1)$. ‖ ‖ step 10 ⋮ Combine results: $\int_{0}^{1} 8t \cosh t \, dt = 8 \sinh 1 - 8 (\cosh 1 - 1) = 8 \sinh 1 - 8 \cosh 1 + 8$. ‖ ∻Answer∻ ⚹ $\int_{0}^{1} 8t \cosh t \, dt = 8 \sinh 1 - 8 \cosh 1 + 8$ ⚹ ∻Key Concept∻ ⚹Integration by parts⚹ ∻Explanation∻ ⚹Integration by parts is used to integrate the product of two functions, and boundary terms must be evaluated for definite integrals.⚹◊What is the general formula for integration by parts in calculus⍭ Generate me a similar question◊

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