Chapter 13: Problem 3
Determine: (a) \(\int 2 \sin (3 x+1) \mathrm{d} x\) (e) \(\int 4^{2 x-3} \mathrm{~d} x\) (b) \(\int \sqrt{5-2 x} \mathrm{~d} x\) (f) \(\int 6 \cos (1-2 x) \mathrm{d} x\) (c) \(\int 6 . e^{1-3 x} \mathrm{~d} x\) (g) \(\int \frac{5}{3 x-2} d x\) (d) \(\int(4 x+1)^{3} \mathrm{~d} x\) (h) \(\int 3 \sec ^{2}(1+4 x) d x\)
Short Answer
Step by step solution
Integrate \( \int 2 \sin (3x+1) \, \mathrm{d}x \)
Integrate \( \int 4^{2x-3} \, \mathrm{d}x \)
Integrate \( \int \sqrt{5-2x} \, \mathrm{d}x \)
Integrate \( \int 6 \cdot e^{1-3x} \, \mathrm{d}x \)
Integrate \( \int(4x+1)^3 \, \mathrm{d}x \)
Integrate \( \int 6 \cos (1-2x) \, \mathrm{d}x \)
Integrate \( \int \frac{5}{3x-2} \, \mathrm{d}x \)
Integrate \( \int 3 \sec^2(1+4x) \, \mathrm{d}x \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
- Pick a substitution, such as \( u = g(x) \), where \( g(x) \) is a part of the integrand.
- Find the derivative \( \frac{\mathrm{d}u}{\mathrm{d}x} \) to express \( \mathrm{d}x \) in terms of \( \mathrm{d}u \).
- Rewrite the integral entirely in terms of \( u \) and \( \mathrm{d}u \).
- Integrate the new, simpler function with respect to \( u \).
- Substitute back the original variable \( x \) by replacing \( u \) again with \( g(x) \).
Indefinite Integrals
- The integral does not have bounds, meaning it yields a general form of the function.
- The constant \( C \) is essential because integration rules allow the addition of any constant to \( F(x) \) without affecting its derivative \( f(x) \).
- For polynomial expressions, when you integrate \( x^n \), the result is \( \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).
- Special functions like sin and cos have specific antiderivatives (-cos for sin and sin for cos, respectively).
Exponential Functions
- Form the expression as \( e^{ax} \), where \( a \) is a constant.
- Utilize the property that the integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} + C \).
Trigonometric Integrals
- The integral \( \int \sin(u) \mathrm{d}u \) results in \(-\cos(u) + C \).
- Similarly, \( \int \cos(u) \mathrm{d}u \) leads to \( \sin(u) + C \).