Question

3.) a.) Evaluate the following integral. (15 points) in(1 +5x?) dx b.) Evaluate the following integral. (10 points) tanº (6x) sec 10 (6x) dx

Answer 1

3. (a) Sen(1+522) di In (1+52) de-ten (14527 dx?da lox x ln (1+5x²) x da 1+5x² = a dn (1+50²) - 10 x² 145x2 da = x ln (1+587) - 2. s 1+5*²1 (1 + 5*2) dan. da. ln (1452) - 2 [ J de Jit's 1 +502 alu (1+5x²) – 20+ 2. 1+5x²

x ln (1+58²) - 2x - 2 op/N 5 da + x² x ln (1+5x2) - 2x + 2/ 2 5 tan 6%) 40 (755) c = Integrating Constant. = n ln (1+5*²) - 2x + 2 & tan-" (55x) + c = x ln (1+5*²) – 2x + 2 tan ! (55x) + c 2x + Sdn (1+5*²) dx = x ln (1+5*²) - 2 2 tan! (25x) + c where, ca Integrating Integrating constant

3.(6) Let 6x = Z 6 dx = dz stmi (52sec" cez de fton perdez de de & Stanoz. sedz. seez da →dx=Ź dz 4. Νσω, secz = 1+ tanz → 8ec²zA= (1+ tantz) secoz = 1+ 4 tanz+6 tantz + 4 tane at taniz, y tan²z. secoz a tan3z (1+ 4 tantz + 6 tamiz + 4 tane x + tan82) = tanz + 4 tan5z + 6 tantz+ 4 tamiz + tan"z

Hence, itong tam3 (6x). seco (6x) da (tan?z. z. Sec&z). sec z dz (tan? Ż+ 4 tandz +6 tantz + 4 tan'z + tan"z) sec z dz 6 - Sairaurburu u + but + 40 + "") du 4. 6 48. ulo u 4 Let tamZ 6. + 4. 6 8 u 12 secz dz=du of C 12 C - T Integrating Constanta constant, 4 - 8 u + -| u + и I u 12 24 9 8 + c 15 72 4 + 24 (tan Z) (tan 2) + I (tanz) (tam zlo I tanz) "? 9 + + + c 15 72

I tan (62) + 4 24 1 tane (6x) + 1 tane (6x) I tan 26x) & C I tan (6x) + t 72 Hence, dan? (6x) seco (6x) da I tano (6x)+ I tant (6x) + 4 I tan (6) + I tanto (6x) + 1/ tan² (6x) .tc, c= Integrating 8 24 15 Constant