Special Considerations•As with differentiation, integration (definite and indefinite) requires significant simplification - both before and afterthe actual calculus (integration) takes place.8
–Rules of Exponents–Trigonometric properties–Algebraic techniques (e.g. solving forx-intercepts)–Arithmetic - this is sometimes the hardest part!•Integration by substitution (Section 5.5) can be quite difficult - especially definite integration by substitution. There isno ‘one size fits all’ approach to these types of integrals. Practice is the only way to become proficient. Be sure to workadditional 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 willfrequently evaluate expressions like (a-b)-(c-d).If you do not use parentheses around the second half of thatexpression 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 withtrigonometric integrands. For instance:Zsinxcosxdx=12sin2x+C=-12cos2x+C=-14cos 2x+C-2π-3π2-π-π2π2π3π22π-11yxf(x) =12sin2xf(x) =-12cos2xf(x) =-14cosxA quick inspection of the graphs of all three solutions will easily demonstrate that they are indeed vertical translationsof each other - identical otherwise (see above).This can also be verified algebraically using the Pythagorean anddouble-angle identities.Consider alsoZ14xdx=14ln|4x|+C= ln|x|+CThe first solution is reached by applying the basic substitutionu= 4xwhereas the second solution is obtained afterfactoring14out of the integrand at the onset and performing basic integration afterwards.The equivalency of thesolutions can be seen by applying the properties of logarithms. Namely,ln|4x|+C= (ln 4|x|) +C= ln 4 + ln|x|+C= ln|x|+Csince ln 4 +Cis indeed itself a constant.9
Sample ExercisesFind the indefinite integral and check the result by differentiation. (Section 5.1)1.ZdxF(x) =x+C2.Z(et-csc2t+ csctcott)dtF(x) =et+ cott-csct+C3.Z-5dxF(x) =-5x+C4.Z2x3+13x2-2x+ 3dxF(x) =12x4+19x3-x2+ 3x+C5.Zx+ 6√xdxF(x) =23x12(x+ 18) +C6.Z(sec2y+ 1)dyF(x) = tany+y+C7.Z(5 cosx+ 4 sinx)dxF(x) = 5 sinx-4 cosx+C8.Z(x+ 1)(3x-2)dxF(x) =x3+12x2-2x+C9.Z3√x5+x-3dxF(x) =38x83-12x2+C10.Z3xdxF(x) = 3 ln|x|+C11.Zdx4x2F(x) =-14x+C12.Zx4-3x2+ 5x4dxF(x) =x+3x-53x3+C13.Z(4t2+ 3)2dtF(x) =16t55+ 8t3+ 9t+C14.Z(cosx+ 3x)dxF(x) = sinx+3xln 3+C15.Z√x+12√xdxF(x) =23x3/2+x1/2+C16.Z(2 sinx-5ex)dxF(x) =-2 cosx-5ex+C17.