Review Of Integration Of Various

Q1: ∫8e^(5x) dx u = 5x du = 5 dx = 8/5 ∫5e^(5x) dx = 8/5 ∫e^u du = 8/5 e^u + C = 8/5 e^(5x) + C Q2: ∫(2e^(2/x⁵))/x⁶ dx u = 2/x⁵ du = -10/x⁶ dx = -1/5 ∫(-10e^(2/x⁵))/x⁶ dx = -1/5 ∫e^u du = -1/5 e^u + C = -1/5 e^(2/x⁵) + C Q3: ∫-2e^(2cos(5x)) sin(5x)dx u = 2cos(5x) du = -10sin(5x) dx = 1/5 ∫-10e^(2cos(5x)) sin(5x) dx = 1/5 ∫e^u du = 1/5 e^u + C = 1/5 e^(2cos(5x)) + C Q4: ∫-5e^(2x) csc²(e^(2x)) dx u = e^(2x) du = 2e^(2x) dx = -5/2 ∫2e^(2x) csc²(e^(2x)) dx = -5/2 ∫csc²(u) du = -5/2 (-cot(u)) + C = 5/2 cot(e^(2x)) + C Q5: ∫(6x²+x)/(4x³+x²) dx u = 4x³+x² du = 12x²+2x dx = 2(6x²+x) dx = 1/2 ∫(12x²+2x)/(4x³+x²) dx = 1/2 ∫(du/u) = 1/2 ln|u| + c = 1/2 ln|4x³+x²| + C The Integral of a Natural logarithmic Function: ∫ln(x)dx = xln(x) - x + C = x(ln(x) - 1) + C Q6: ∫6ln(x)dx = 6∫ln(x)dx = 6xln(x) - 6x + C Q7: ∫-2tan(9x)dx = ∫(-2sin(9x)/cos(9x)) dx u = cos(9x) du = -sin(9x)(9)dx = 2/9 ∫(-9sin(9x)/cos(9x)) dx = 2/9 ∫(du/u) = 2/9 ln|u| + C = 2/9 ln|cos(9x)| + C Q8: ∫(-4e^(√4x))/√x dx u = √4x = 2x¹ᐟ² du = x⁻¹ᐟ² = 1/√x dx = -2∫(e^(√4x)/√x) dx = -2∫e^u du = -2e^u + C = -2e^(√4x) + C Q9: ∫-5e^(2x) csc²(e^(2x)) dx u = e^(2x) du = 2e^(2x) dx = -5/2 ∫2e^(2x) csc²(e^(2x)) dx = -5/2 ∫csc²(u) du = -5/2 [-cot(u)] + C = -5/2 [-cot(e^(2x))] + C Q10: ∫5tan(2x)ln(cos(2x))dx u = ln(cos(2x)) du = (-2sin(2x)/cos(2x)) dx = -2tan(2x)dx = -5/2 ∫-2tan(2x)ln(cos(2x))dx = -5/2 ∫u du = -5/2 * u²/2 + C = -5/4 [ln(cos(2x))]² + C Integration Formula Resulting In An Inverse Trigonometric Function: ∫(1/√(a²-u²))du = sin⁻¹(u/a) + C Q11: ∫(ln(-2x))/(5x) dx u = ln(-2x) du = 1/x dx = 1/5 ∫(ln(-2x)/x) dx = 1/5 ∫u du = 1/5 * 1/2 u² + C = 1/10 [ln(-2x)]² + C Q12: ∫8/√(49-16x²) dx u²=16x² -> u=4x du=4dx -> dx=1/4 du = ∫8/√(49-16x²) dx = ∫2/√(49-u²) du = 2∫1/√(49-u²) du = 2∫1/√(7²-u²) du = 2sin⁻¹(u/7) + C = 2sin⁻¹(4x/7) + C Rule: Integration Formulas Resulting in Inverse Trigonometric Functions The following integration formulas yield inverse trigonometric functions: ∫(1/√(a²-u²))du = sin⁻¹(u/a) + C ∫(1/(a²+u²))du = (1/a)tan⁻¹(u/a) + C ∫(1/(u√(u²-a²)))du = (1/a)sec⁻¹(u/a) + C Q13: ∫1/(16+16x²) dx u²=16x² -> u=4x du=4dx -> dx=1/4 du = ∫1/(4(16+u²)) du = 1/4 ∫1/(16+u²) du = 1/4 * 1/4 tan⁻¹(u/4) + C = 1/16 tan⁻¹(4x/4) + C = 1/16 tan⁻¹(x) + C Q14: ∫(from -2 to 2) 3/(4+9x²) dx u²=4x² -> u=2x du=2dx -> dx=1/2 du = ∫(from -2 to 2) 3/(2(4+u²)) du = 3/2 ∫(from -2 to 2) 1/(4+u²) du = 3/2 * 1/2 tan⁻¹(u/2) | (from -2 to 2) = [3/4 tan⁻¹(2/2)] - [3/4 tan⁻¹(-2/2)] = (3π/16) - (-3π/16) = 3π/8 Q15: ∫(from -2√3/3 to 0) 11/(4+4x²) dx u²=4x² -> u=2x du=2dx -> dx=1/2 du = ∫(from -2√3/3 to 0) 11/(2(4+u²)) du = 11/2 ∫(from -2√3/3 to 0) 1/(4+u²) du = 11/2 * 1/2 tan⁻¹(u/2) | (from -2√3/3 to 0) = [11/4 tan⁻¹(0/2)] - [11/4 tan⁻¹((-2√3/3)/2)] = 11π/24 Q16: ∫(-12sec⁻¹(3x))/(x√(9x²-1)) dx u = sec⁻¹(3x) du = 1/(x√(9x²-1)) dx = -12 ∫u du = -12 u²/2 + C = -6u² + C = -6[sec⁻¹(3x)]² + C Q17: ∫9/(x√(49x²-4)) dx u²=49x² -> u=7x du=7dx -> dx=1/7 du = ∫(9/( (u/7)√(u²-4) )) (1/7)du = ∫9/(u√(u²-4)) du = 9∫1/(u√(u²-4)) du = 9∫1/(u√(u²-2²)) du = 9 * 1/2 sec⁻¹(u/2) + C = 9/2 sec⁻¹(7x/2) + C Q18: ∫(from -2/5 to 0) 14/√(25-25x²) dx u²=25x² -> u=5x du=5dx -> dx=1/5 du = ∫(from -2/5 to 0) 14/(5√(25-u²)) du = 14/5 ∫(from -2 to 0) 1/√(25-u²) du = 14/5 [sin⁻¹(u/5)] | (from -2 to 0) = [14/5 sin⁻¹(0/5)] - [14/5 sin⁻¹(-2/5)] = 0 - (-7π/15) = 7π/15 Q19: ∫(from 2√3/3 to 2) 12/(x√(9x²-9)) dx u²=9x² -> u=3x -> x=u/3 du=3dx -> dx=1/3 du = ∫(from 2√3 to 6) 12 / ( (u/3)√(u²-9) ) (1/3) du = ∫(from 2√3 to 6) 12/(u√(u²-9)) du = 12∫(from 2√3 to 6) 1/(u√(u²-9)) du = 12 * [1/3 sec⁻¹(u/3)] | (from 2√3 to 6) = [4sec⁻¹(6/3)] - [4sec⁻¹(2√3/3)] = 4sec⁻¹(2) - 4sec⁻¹(2√3/3) = 4π/3 - 2π/3 = 2π/3