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yuanzhi's Question
Math
Posted 22 days ago

3. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.1.011.MI.
ASK YOUR TEACHER
PRACTICE ANOTHER
Evaluate the integral. (Use CC for the constant of integration.)
arctan6tdt\int \arctan 6 t d t
\square
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4. [/1[-/ 1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.1.013.MI.
ASK YOUR TEACHER
PRACTICE ANOTHER
Evaluate the integral. (Use CC for the constant of integration.)
e2θsin(3θ)dθ
\int e^{2 \theta} \sin (3 \theta) d \theta

\square
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5. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.1.021.
ASK YOUR TEACHER
PRACTICE ANOTHER
Evaluate the integral.
018tcoshtdt
\int_{0}^{1} 8 t \cosh t d t

\square
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Answer from Sia
Posted 22 days ago
Solution by Steps
step 1
Let I=arctan6tdtI = \int \arctan 6t \, dt
step 2
Use integration by parts: Let u=arctan6tu = \arctan 6t and dv=dtdv = dt
step 3
Compute dudu and vv: du=61+(6t)2dtdu = \frac{6}{1 + (6t)^2} \, dt and v=tv = t
step 4
Apply the integration by parts formula: udv=uvvdu\int u \, dv = uv - \int v \, du
step 5
Substitute uu, vv, dudu, and dvdv into the formula: arctan6tdt=tarctan6tt61+36t2dt\int \arctan 6t \, dt = t \arctan 6t - \int t \cdot \frac{6}{1 + 36t^2} \, dt
step 6
Simplify the integral: t61+36t2dt=6t1+36t2dt\int t \cdot \frac{6}{1 + 36t^2} \, dt = \int \frac{6t}{1 + 36t^2} \, dt
step 7
Use substitution: Let u=36t2+1u = 36t^2 + 1, then du=72tdtdu = 72t \, dt
step 8
Rewrite the integral: 6t1+36t2dt=6tudu72t=1121udu\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: 1121udu=112lnu+C=112ln36t2+1+C\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: arctan6tdt=tarctan6t112ln36t2+1+C\int \arctan 6t \, dt = t \arctan 6t - \frac{1}{12} \ln |36t^2 + 1| + C
Answer
arctan6tdt=tarctan6t112ln36t2+1+C\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 e2θsin(3θ)dθ\int e^{2\theta} \sin(3\theta) \, d\theta
step 1
Let I=e2θsin(3θ)dθI = \int e^{2\theta} \sin(3\theta) \, d\theta
step 2
Use integration by parts twice: Let u=e2θu = e^{2\theta} and dv=sin(3θ)dθdv = \sin(3\theta) \, d\theta
step 3
Compute dudu and vv: du=2e2θdθdu = 2e^{2\theta} \, d\theta and v=13cos(3θ)v = -\frac{1}{3} \cos(3\theta)
step 4
Apply the integration by parts formula: udv=uvvdu\int u \, dv = uv - \int v \, du
step 5
Substitute uu, vv, dudu, and dvdv into the formula: e2θsin(3θ)dθ=13e2θcos(3θ)13cos(3θ)2e2θdθ\int e^{2\theta} \sin(3\theta) \, d\theta = -\frac{1}{3} e^{2\theta} \cos(3\theta) - \int -\frac{1}{3} \cos(3\theta) \cdot 2e^{2\theta} \, d\theta
step 6
Simplify the integral: e2θsin(3θ)dθ=13e2θcos(3θ)+23e2θcos(3θ)dθ\int e^{2\theta} \sin(3\theta) \, d\theta = -\frac{1}{3} e^{2\theta} \cos(3\theta) + \frac{2}{3} \int e^{2\theta} \cos(3\theta) \, d\theta
step 7
Use integration by parts again: Let u=e2θu = e^{2\theta} and dv=cos(3θ)dθdv = \cos(3\theta) \, d\theta
step 8
Compute dudu and vv: du=2e2θdθdu = 2e^{2\theta} \, d\theta and v=13sin(3θ)v = \frac{1}{3} \sin(3\theta)
step 9
Apply the integration by parts formula: udv=uvvdu\int u \, dv = uv - \int v \, du
step 10
Substitute uu, vv, dudu, and dvdv into the formula: e2θcos(3θ)dθ=13e2θsin(3θ)13sin(3θ)2e2θdθ\int e^{2\theta} \cos(3\theta) \, d\theta = \frac{1}{3} e^{2\theta} \sin(3\theta) - \int \frac{1}{3} \sin(3\theta) \cdot 2e^{2\theta} \, d\theta
step 11
Simplify the integral: e2θcos(3θ)dθ=13e2θsin(3θ)23e2θsin(3θ)dθ\int e^{2\theta} \cos(3\theta) \, d\theta = \frac{1}{3} e^{2\theta} \sin(3\theta) - \frac{2}{3} \int e^{2\theta} \sin(3\theta) \, d\theta
step 12
Combine results: e2θsin(3θ)dθ=13e2θcos(3θ)+23(13e2θsin(3θ)23e2θsin(3θ)dθ)\int e^{2\theta} \sin(3\theta) \, d\theta = -\frac{1}{3} e^{2\theta} \cos(3\theta) + \frac{2}{3} \left( \frac{1}{3} e^{2\theta} \sin(3\theta) - \frac{2}{3} \int e^{2\theta} \sin(3\theta) \, d\theta \right)
step 13
Solve for the integral: I=13e2θcos(3θ)+29e2θsin(3θ)49II = -\frac{1}{3} e^{2\theta} \cos(3\theta) + \frac{2}{9} e^{2\theta} \sin(3\theta) - \frac{4}{9} I
step 14
Combine like terms: I+49I=13e2θcos(3θ)+29e2θsin(3θ)I + \frac{4}{9} I = -\frac{1}{3} e^{2\theta} \cos(3\theta) + \frac{2}{9} e^{2\theta} \sin(3\theta)
step 15
Simplify: 139I=13e2θcos(3θ)+29e2θsin(3θ)\frac{13}{9} I = -\frac{1}{3} e^{2\theta} \cos(3\theta) + \frac{2}{9} e^{2\theta} \sin(3\theta)
step 16
Solve for II: I=913(13e2θcos(3θ)+29e2θsin(3θ))+CI = \frac{9}{13} \left( -\frac{1}{3} e^{2\theta} \cos(3\theta) + \frac{2}{9} e^{2\theta} \sin(3\theta) \right) + C
step 17
Simplify: I=313e2θcos(3θ)+213e2θsin(3θ)+CI = -\frac{3}{13} e^{2\theta} \cos(3\theta) + \frac{2}{13} e^{2\theta} \sin(3\theta) + C
Answer
e2θsin(3θ)dθ=313e2θcos(3θ)+213e2θsin(3θ)+C\int e^{2\theta} \sin(3\theta) \, d\theta = -\frac{3}{13} e^{2\theta} \cos(3\theta) + \frac{2}{13} e^{2\theta} \sin(3\theta) + 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 018tcoshtdt\int_{0}^{1} 8t \cosh t \, dt
step 1
Let I=018tcoshtdtI = \int_{0}^{1} 8t \cosh t \, dt
step 2
Use integration by parts: Let u=8tu = 8t and dv=coshtdtdv = \cosh t \, dt
step 3
Compute dudu and vv: du=8dtdu = 8 \, dt and v=sinhtv = \sinh t
step 4
Apply the integration by parts formula: udv=uvvdu\int u \, dv = uv - \int v \, du
step 5
Substitute uu, vv, dudu, and dvdv into the formula: 018tcoshtdt=[8tsinht]01018sinhtdt\int_{0}^{1} 8t \cosh t \, dt = \left[ 8t \sinh t \right]_{0}^{1} - \int_{0}^{1} 8 \sinh t \, dt
step 6
Evaluate the boundary term: [8tsinht]01=8sinh10=8sinh1\left[ 8t \sinh t \right]_{0}^{1} = 8 \sinh 1 - 0 = 8 \sinh 1
step 7
Simplify the integral: 018sinhtdt=801sinhtdt\int_{0}^{1} 8 \sinh t \, dt = 8 \int_{0}^{1} \sinh t \, dt
step 8
Integrate: sinhtdt=cosht\int \sinh t \, dt = \cosh t
step 9
Evaluate the boundary term: 8[cosht]01=8(cosh1cosh0)=8(cosh11)8 \left[ \cosh t \right]_{0}^{1} = 8 (\cosh 1 - \cosh 0) = 8 (\cosh 1 - 1)
step 10
Combine results: 018tcoshtdt=8sinh18(cosh11)=8sinh18cosh1+8\int_{0}^{1} 8t \cosh t \, dt = 8 \sinh 1 - 8 (\cosh 1 - 1) = 8 \sinh 1 - 8 \cosh 1 + 8
Answer
018tcoshtdt=8sinh18cosh1+8\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.

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