Special considerations as with differentiation

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Special Considerations As with differentiation, integration (definite and indefinite) requires significant simplification - both before and after the actual calculus (integration) takes place. 8
Rules of Exponents Trigonometric properties Algebraic techniques (e.g. solving for x -intercepts) Arithmetic - this is sometimes the hardest part! Integration by substitution (Section 5.5) can be quite difficult - especially definite integration by substitution. There is no ‘one size fits all’ approach to these types of integrals. Practice is the only way to become proficient. Be sure to work additional exercises beyond the graded homework. Work on this study guide as well as exercises from the textbook. Be liberal with your use of parentheses, especially when using the first Fundamental Theorem of Calculus. You will frequently evaluate expressions like ( a - b ) - ( c - d ). If you do not use parentheses around the second half of that expression and then distribute the negative sign, you will end up with the wrong solution! As mentioned before, antiderivatives are non-unique (without additional initial conditions). Due to this fact, occasion- ally one may reach two (or more!) ostensibly different solutions for a single antiderivative. This often happens with trigonometric integrands. For instance: Z sin x cos xdx = 1 2 sin 2 x + C = - 1 2 cos 2 x + C = - 1 4 cos 2 x + C - 2 π - 3 π 2 - π - π 2 π 2 π 3 π 2 2 π - 1 1 y x f ( x ) = 1 2 sin 2 x f ( x ) = - 1 2 cos 2 x f ( x ) = - 1 4 cos x A quick inspection of the graphs of all three solutions will easily demonstrate that they are indeed vertical translations of each other - identical otherwise (see above). This can also be verified algebraically using the Pythagorean and double-angle identities. Consider also Z 1 4 x dx = 1 4 ln | 4 x | + C = ln | x | + C The first solution is reached by applying the basic substitution u = 4 x whereas the second solution is obtained after factoring 1 4 out of the integrand at the onset and performing basic integration afterwards. The equivalency of the solutions can be seen by applying the properties of logarithms. Namely, ln | 4 x | + C = (ln 4 | x | ) + C = ln 4 + ln | x | + C = ln | x | + C since ln 4 + C is indeed itself a constant. 9
Sample Exercises Find the indefinite integral and check the result by differentiation. (Section 5.1) 1. Z dx F ( x ) = x + C 2. Z ( e t - csc 2 t + csc t cot t ) dt F ( x ) = e t + cot t - csc t + C 3. Z - 5 dx F ( x ) = - 5 x + C 4. Z 2 x 3 + 1 3 x 2 - 2 x + 3 dx F ( x ) = 1 2 x 4 + 1 9 x 3 - x 2 + 3 x + C 5. Z x + 6 x dx F ( x ) = 2 3 x 1 2 ( x + 18) + C 6. Z ( sec 2 y + 1 ) dy F ( x ) = tan y + y + C 7. Z (5 cos x + 4 sin x ) dx F ( x ) = 5 sin x - 4 cos x + C 8. Z ( x + 1)(3 x - 2) dx F ( x ) = x 3 + 1 2 x 2 - 2 x + C 9. Z 3 x 5 + x - 3 dx F ( x ) = 3 8 x 8 3 - 1 2 x 2 + C 10. Z 3 x dx F ( x ) = 3 ln | x | + C 11. Z dx 4 x 2 F ( x ) = - 1 4 x + C 12. Z x 4 - 3 x 2 + 5 x 4 dx F ( x ) = x + 3 x - 5 3 x 3 + C 13. Z ( 4 t 2 + 3 ) 2 dt F ( x ) = 16 t 5 5 + 8 t 3 + 9 t + C 14. Z (cos x + 3 x ) dx F ( x ) = sin x + 3 x ln 3 + C 15. Z x + 1 2 x dx F ( x ) = 2 3 x 3 / 2 + x 1 / 2 + C 16. Z (2 sin x - 5 e x ) dx F ( x ) = - 2 cos x - 5 e x + C 17.

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