Title 73 Direct Integration.pdf ✅

3. Exponential Functions [DERIVATION] From differentiation, d dx (a x ) = a x ln a Therefore, its antiderivative is ∫ a xdx = 1 ln a a x + C Since when a = e, ln e = 1 ∫ e xdx = e x + C 4. Trigonometric Functions From inverse differentiation, the integrals of sin x and cos x result to ∫ sin x dx = − cosx + C ∫ cos x dx = sin x + C For tan x, consider quotient identities ∫ tan x dx = ∫ sin x cos x dx = − ∫ d(cos x) cos x = − ln|cos x| + C ∫ tan x dx = ln|sec x| + C Similarly, ∫ cot x dx = ln|csc x| + C For sec x, its integration is a challenge. ∫ sec x dx = ∫ sec x sec x + tan x sec x + tan x dx The multiplier may be non-intuitive, but it is just a multiplier equivalent to 1, which means it does not affect the value. = ∫ sec2 x + sec x tan x sec x + tan x dx = ∫ d(sec x + tan x) sec x + tan x ∫ sec x dx = ln | sec x + tan x | + C
Similarly, ∫ csc x dx = − ln|csc x + cot x| + C 5. Hyperbolic Functions The definitions of each hyperbolic function are: sinh x = e x − e −x 2 cosh x = e x + e −x 2 tanh x = sinh x cosh x csch x = 1 sinh x sech x = 1 cosh x coth x = 1 tanh x From the definitions, one may derive the integral formulas of such functions ∫ sinh x dx = cosh x + C ∫ cosh xdx = sinh x + C ∫ tanh x dx = ln|cosh x| + C ∫ csch x dx = ln|coth x − csch x| + C ∫ sech x dx = arctan(sinh x) + C ∫ coth x dx = ln | sinh x | + C 6. Integrals that Result in Logarithms One can recall that d dx (ln x) = 1 x Therefore, manipulating this results to ∫ du u = ln |u| + C - Example: ∫ x x 2+1 dx [SOLUTION] ∫ x x 2 + 1 dx = 1 2 ∫ 2x x 2 + 1 dx