Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
please answer number 4 and follow the format of sample solution (include the rules/techniques used)
![Evaluate the following integrals. Use only the variables "u" (and "v" if applicable)
in your substitutions.
1. S
dx
Vx+2
2. S
dx
3. ſesx tan e 5x dx
4. S(x² – 1) Vx + 4 dx
5. S dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F888382d9-fd0a-4ffd-a6f8-eb9e9a55ed67%2F6ccbe99e-060a-433e-b597-c6230289dd57%2Fkkspq1a_processed.png&w=3840&q=75)
Transcribed Image Text:Evaluate the following integrals. Use only the variables "u" (and "v" if applicable)
in your substitutions.
1. S
dx
Vx+2
2. S
dx
3. ſesx tan e 5x dx
4. S(x² – 1) Vx + 4 dx
5. S dx
![Sample solution with steps:
Evaluate the integral.
3-2 sin x
dx
3x+2 cos x
Step 1: Rewrite the equation in the form f(u)du.
Let u = 3x + 2 cos x
du = (3 – 2sin x) dx
Substitute u and du on the original equation.
3-2 sinx
3x+2 cos x
dx = J (+2 cos x
) (3 – 2 sin x)dx
- SC) du
Step 2: Integrate S () du.
Use the integration rule: S = In x + C
S) du = In u + C
Step 3: Substitute back u = 3x + 2 cos x.
In u + C = In (3x + 2 cos x) + C
3-2 sin x
Therefore, S
dx = In (3x +2 cos x) + C
3x+2 cos x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F888382d9-fd0a-4ffd-a6f8-eb9e9a55ed67%2F6ccbe99e-060a-433e-b597-c6230289dd57%2Fqdoduji_processed.png&w=3840&q=75)
Transcribed Image Text:Sample solution with steps:
Evaluate the integral.
3-2 sin x
dx
3x+2 cos x
Step 1: Rewrite the equation in the form f(u)du.
Let u = 3x + 2 cos x
du = (3 – 2sin x) dx
Substitute u and du on the original equation.
3-2 sinx
3x+2 cos x
dx = J (+2 cos x
) (3 – 2 sin x)dx
- SC) du
Step 2: Integrate S () du.
Use the integration rule: S = In x + C
S) du = In u + C
Step 3: Substitute back u = 3x + 2 cos x.
In u + C = In (3x + 2 cos x) + C
3-2 sin x
Therefore, S
dx = In (3x +2 cos x) + C
3x+2 cos x
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