u C = 1 18 (2x3 + 5) 3 + C This is the answer, which we can check by di erentiating: = (2x3 + 5) 2 (6x 2 )+0=x 2 (2x 3 + 5) 2

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2 Net multiply by, bring the unwanted outside the integral, and substitute in. ( + 5) d = ( + 5) {z {z d = u u We can now use Formula getting u = u + C = 8 ( + 5) + C This is the answer, which we can check by di erentiating: apple 8 ( + 5) + C 0 = 8 ( + 5) ( )+0= ( + 5) which is the original integrand, verifying that our answer is correct. sin Eample. Find [5 + cos ] d. Solution. Again we have a quantity to a power, so we let u equal the quantity and try to get. u = 5 + cos, = sind We work towards putting this in the form u n. To avoid unnecessary confusion, we move the outside the integral and rewrite this as sin d = [5 + cos ] [5 + cos ] sin d We proceed as before: We use Formula and get = [5 + cos ] = sin d [5 + cos ] ( sind) = {z {z u u = u + C = 9 [5 + cos ] + C = 9(5 + cos ) + C This is the answer which you check by di erentiating 9 [5 + cos ] + C 0 = 9 ( )[5 + cos ] ( sin ) + 0 = sin [5 + cos ]

3 Substitution with Trig Functions. There are four fundamental trig di erentiation formulas: d(sin u) = cos u d(tan u) =sec u d(cos u) = sin u d(sec u) =secu tan u By going backwards, these lead to four fundamental integration formulas. () cos u=sinu + C sin u= cos u + C sec u= tan u + C sec u tan u=secu + C There are similar formulas involving cot u and csc u. Eample. Find sin(5 + )d. Solution. Here, we do not have a quantity to a power, so the first principle of substitution () does not apply. Hence, we turn to the second principle of substitution. () When you do not have a quantity to a power, but you do have a trig function of a quantity let u equal that quantity and try to get. Later, this same principle will equally apply to other kinds of functions, such as eponential, logarithmic, inverse trig, etc. Let u =5 +. Then = 0 d We work towards putting the problem in the form sin u.we first move the net to the d. sin(5 + )d = sin(5 + ) d Net, we multiply by 0, bring the unwanted outside of the integral, and substitute in 0 0 sin(5 + ) 0 0 d = sin (5 + ) 0 0 {z {z d = sin u 0 u

4 We now use the appropriate formula from () getting [ cos u]+c = 0 0 cos(5 + ) + C This is the answer, which we can check by di erentiating apple 0 0 cos(5 + ) + C = sin(5 + )(0 ) +0= sin(5 + ) 0 Eample. Find 9sec ( p + 5) p d. Solution. We recognize that we have a formula for problem into this form. sec uin (), so we move toward putting the Let u = p +5= / + 5. Then = / d = We first move the 9 outside of the integral and move the p d p net to the d. 9sec ( p + 5) p d =9 sec ( p + 5) p d Net multiply by, move the unwanted outside of the integral, simplify and substitute. 9 sec ( p + 5) p d =9 sec ( p + 5) {z p d = sec u u {z We apply the appropriate formula from () getting tan u + C = tan( p + 5) + C This is the answer, which you should check by di erentiating, but we shall not here. Skipping Steps. The first five or so times you use substitution, you should write out all of the steps. Thereafter, you should skip steps that you can easily do in your head and that leaving out does not lead to errors. For instance, in Eample, ( + 5) d, we let u = + 5 and find = d. Many people would not physically move the net to the d, nor would they multiply by, but they would just think these steps. They would proceed by: ( + 5) d = ( + 5) d = u = u + C = 8 ( + 5) + C

5 When you are really eperienced and the problem is simple enough, you might want to try doing more in your head. But be careful, and, again, if the work is being graded, make sure you write down enough so that the grader can follow what you do. The balance is, as it is throughout mathematics, you want to write down enough details so that you do not make errors, but not too many details so that the flow of the problem is disrupted in your mind. Special Case. Very simple problems like cos 7 dshould be handled in a simple way. Remember, you know 7 d = C and not 8 + C. You need the 8 to cancel the 8 when checking (8 ) 0 =8 7. This is a similar situation. You know cos d=sin+c. As with 7 d, cos 7 d= sin 7+C and not sin 7+C, 7 because when you check, (sin 7) 0 = 7 cos 7 and you need the to cancel the 7. Similarly, 7 sec tan d =sec + C since the is needed to cancel the when you check the answer. When Substitution Does Not Work. Sometimes the obvious choice of substitution does not work. This method of substitution allows you to see as soon as possible that it does not work, and you must look for an alternative method. Eample 5. Find ( + 5) d (instead of ( + 5) d of Eample ). Solution. We see we have a quantity to a power, so (as in Eample ) we try, u = +5, = d. But right here we see we cannot get since we only have an. (Warning: We cannot multiply by, since we can bring the outside the integral sign, but we cannot bring the outside.) Thus we must try something else. The thing to do here is to square out the quantity and then multiply through with the. ( + 5) d = ( ) d = ( ) d = Eample. Find 9sec ( p + 5) p d, from Eample. Solution. Suppose we see that we have a quantity to a power, and we try u =sec( p + 5), =sec( p + 5)) tan( p + 5) We can see right away that there is no sec( ) tan( ). Thus, we see we cannot get, so we look for an alternative approach, and hope we see the method of Eample. p 5

6 We conclude with an eample from Calc. II. This will illustrate that we shall learn many methods of integration there, and it is important to see as quickly as possible whether a particular method does or does not work on a given problem. Here are two formulas that will be derived in Calc. II. (5) u = `n u + C, +u = tan u + C The first formula is interesting because it tells us what happens in Formula when n =. Eample 7. Find (a) d, (b) + + d. Solution. (a) We have a quantity in the denominator, so we try u =+, = d We see with the in the numerator we can get, and we proceed to do it. + d = + {z {z d = u u = `n u + C = `n( + )+C (b) If we were to try u =+ again, we see with only an in the numerator, we cannot get = d. In searching for an alternate approach, we might ask, how do we make use of the single in the numerator? If we observe is the crucial part of the derivative of, and that in the denominator =( ),thenwe solve the problem with the surprising substitution u =, =d Then, + d = +( ) {z u {z d = +u = tan (u)+c = tan ( )+C This completes the discussion.