Practice Exam (Second Midterm), Math1331. Which of the following functions grows faster as x → ∞ ?f(x) = x , g(x) =1sin−11x22. Evaluate the following integrals(a)Z√tan−1x1 + x2dx.(b)Zdx(x − 2)√x2− 4x − 12.(c)Zexsin x dx3. (a) Rewrite (sinh x+cosh x)5in terms of exponential functions and sim-plify as much as you can.(b) EvaluateZπ / 202 sinh(sin θ) cos θ dθ.4. Compute the derivative of the functiony = sin−1(tan−1(ln x))5. Expand the following quotient by partial fractions:tt4− t2− 21Solutions: 1.) We use l’Hˆopital’s rulelimx→∞f(x)g(x)= limx→∞sin−11x21x= limx→∞−2x31p1−1x4−1x2= limx→∞−2x1q1 −1x4= 0,hence the function g(x) grows faster.2a.) We substitute u = tan−1x , du =11+x2dx and getZ√tan−1x1 + x2dx =Z√udu=23u3/2+ C=23(tan−1x)3/2+ C.2b.) We first complete the squarex2− 4x − 12 = x2− 4x + 4 − 16 = (x − 2)2− 16so thatZdx(x − 2)√x2− 4x − 12=Zdx(x − 2)p(x − 2)2− 16u=x−2 , du=dx=Zduu√u2− 16=14sec−1u4+ C=14sec−1x − 24+ C2c.) We perform integration by parts twice. First, we choose u′(x) = ex, v(x) =sin x so thatZexsin xdx = exsin x −Zexcos xdxThen we choose u′(x) = ex, v(x) = cos x, and we getZexsin xdx = exsin x −excos x −Zex· (−sin x)dx= exsin x − excos x −Zexsin xdx.2Finally,2Zexsin xdx = exsin x − excos xandZexsin xdx =12(exsin x − excos x).3a.) Recalling thatsinh x =ex− e−x2, cosh x =ex+ e−x2we obtain(sinh x + cosh x)5= (ex)5= e5x.3b.) We first substitute u = sin θ, du = cos θ dθ so thatZ2 sinh(sin θ) cos θ dθ = 2Zsinh u du= 2 cosh u + C= 2 cosh(sin θ) + Cand finallyZπ / 202 sinh(sin θ) cos θ dθ = 2 [cosh(sin θ)]|π / 20= 2 (cosh(1) − 1)4.) Recall that(sin−1x)′=1√1 − x2, (tan−1x)′=11 + x2, (ln x)′=1x.Using the chain rule we gety′=1p1 − (tan−1(ln x))2·11 + (ln x)2·1x.5.) The equation t4− t2− 2 = 0 is quadratic in t2, and we find t2= 2 sothat there are two solutions t = ±√2. Dividing t4−t2−2 by (t +√2)(t −√2)we obtain 1 + t2so thatt4− t2− 2 = (1 + t2)(t +√2)(t −√2).This leads to the setuptt4− t2− 2=At +√2+Bt −√2+Ct + D1 + t2.Multiplying both sides with (1 + t2)(t +√2)(t −√2) we get the equationt = A(t −√2)(1 + t2) + B(t +√2)(1 + t2) + (1)(Ct + D)(t −√2)(t +√2).3Inserting t =√2 we get√2 = B · 2√2 · 3 = 6√2BandB =16.Inserting t = −√2 we obtain−√2 = A(−2√2) · 3andA =16.We now insert t = 0 and the above values for A, B into equation (1) so that0 = −√26+√26− 2DandD = 0.Inserting A, B, D into equation (1) we obtaint =16(1 + t2)(t −√2 + t +√2) + Ct(t −√2)(t +√2)=13t(1 + t2) + Ct(t2− 2)=13t +13t3+ Ct3− 2Ct=13+ Ct3+13− 2Ctwhich leads toC = −13.Summarizing,tt4− t2− 2=161t +√2+1t −√2−t3(1 + t2)=t3(t2− 2)−t3(1 +