10/8/2021

## Review Usubstitutionap Calculus

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Instructor What we're going to do in this video is get some practice applying u-substitution to definite integrals. So let's say we have the integral, so we're gonna go from x equals one to x equals two, and the integral is two x times x squared plus one to the third power dx. THE METHOD OF U-SUBSTITUTION The following problems involve the method of u-substitution. It is a method for finding antiderivatives. We will assume knowledge of the following well-known, basic indefinite integral formulas.

### Review Usubstitutionap Calculus 3

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### Section 1-3 : Trig Substitutions

For problems 1 – 8 use a trig substitution to eliminate the root.

### Review Usubstitutionap Calculus 2

1. (sqrt {4 - 9{z^2}} ) Solution
2. (sqrt {13 + 25{x^2}} ) Solution
3. ({left( {7{t^2} - 3} right)^{frac{5}{2}}}) Solution
4. (sqrt {{{left( {w + 3} right)}^2} - 100} ) Solution
5. (sqrt {4{{left( {9t - 5} right)}^2} + 1} ) Solution
6. (sqrt {1 - 4z - 2{z^2}} ) Solution
7. ({left( {{x^2} - 8x + 21} right)^{frac{3}{2}}}) Solution
8. (sqrt {{{bf{e}}^{8x}} - 9} ) Solution

### Review Usubstitutionap Calculus Pdf

For problems 9 – 16 use a trig substitution to evaluate the given integral.

1. ( displaystyle int{{frac{{sqrt {{x^2} + 16} }}{{{x^4}}},dx}}) Solution
2. ( displaystyle int{{sqrt {1 - 7{w^2}} ,dw}}) Solution
3. ( displaystyle int{{{t^3}{{left( {3{t^2} - 4} right)}^{frac{5}{2}}},dt}}) Solution
4. ( displaystyle int_{{ - 7}}^{{ - 5}}{{frac{2}{{{y^4}sqrt {{y^2} - 25} }},dy}}) Solution
5. ( displaystyle int_{1}^{4}{{2{z^5}sqrt {2 + 9{z^2}} ,dz}}) Solution
6. ( displaystyle int{{frac{1}{{sqrt {9{x^2} - 36x + 37} }},dx}}) Solution
7. ( displaystyle int{{frac{{{{left( {z + 3} right)}^5}}}{{{{left( {40 - 6z - {z^2}} right)}^{frac{3}{2}}}}},dz}}) Solution
8. ( displaystyle int{{cos left( x right)sqrt {9 + 25 sin^2left( x right)} ,dx}}) Solution
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