Math 132 – Practice Problems for Section 5.5I. Without substitution.1.Z1x2dx2.Z3√5x dx3.Z(1 + x)2dx4.Zsec 3x tan 3x dx5.ddxZx42sin54t dt.6.ddwZ√wwsin54ss4ds.7.ddrZrr2cos 2y tan4y dy.II. Indefinite integrals by substitution. Hint: Set u = what’s inside√or inside ( ).ExampleZxp1 + x2dx.Solution: Set u = 1 + x2. Then du = 2x dx and12du = x dx, andZ21xp1 + x2dx =Z√u12du =12Zu1/2du =12·23u3/2=13(1 + x2)3/28.Zx(2 + x2)3dx9.Z4x2dx√x3+ 810.Zt38 + t4dt11.Z2z + 3√z2+ 3zdzIII. Definite integrals by substitution – Method 1.Example I =Z21x(2x2− 1)3dx.Solution: Set u = 2x2− 1. Then du = 4x dx and14du = x dx, and the integral becomes14Zduu3=14Zu−3du =14u−2−2= −18u2Substitute back and put in the limits of integration:I = −18(2x2− 1)221= −18172−112=−181 − 4949=649.12.Zπ/40sin θcos2θdθ13.Zπ/40sec2t√tan t dtIV. Definite integrals by substitution – Method 2.It is quicker to change the limits of integration to the u variable:Example I =Z10x2(1 + 2x3)5dx.Solution: Set u = 1+2x3. Then du = 6x2dx and16du = x2dx, and the integral becomesI =16Zu(1)u(0)u5du =16u66u(1)u(0)=136u631=136(36− 1).14.Z4−4(y + 4)2/3dy15.Z4−4(y + 4)2/3dy16.Z20x16 +
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