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\fe2x (1-e4x)-2 We will evaluate (sin-1(e2x))cos2(In(sin-1(e2x)) -dx . We can rewrite this by putting term/s with negative exponent in the denominator. e 2x(1 - e4x)-2 (sin 1 (e2x))cos2 (In(sin-1(ezx)) dx e2x -dx (1 - e4x)z(sin-1(e2x)) cos2(In(sin-1(e2x)) 2e2x Let u = In(sin-1(e2x)) and du dx (1-e4x)2(sin-1(e2x)) The derivative of In (sin ](e *)) is obtained by one of the differentiation rules in natural logarithmic functions. [In(u(x))]'= .u'(x); in which u is a function of x. [In (sin 1(ezx) ) ] = sin 1 e2x ' dx (sin 1(ex)) The derivative of sin 1(e2*) is obtained using one of the differentiation rules for inverse trigonometric functions. (Sine function) [sin (u(x))'= u'(x); in which u is a function of x. dx [sin-1(e2x)] = V1 - (e2x)2 dx (ezx) 1 sin -1(ezx) 1 - (e2x)2 dx (ezx) The derivative of e" is obtained by using one of the differentiation rules for exponential functions.1. Evaluate J In 2e* 2arctan ex (1+e2x) sec(2arctan ex ) dx. Use the constant multiple rule: In 2 e * 2arctanex Given that n is a constant: (1 + e2x) sec(2arctan ex) dx In f(x)dx = n [f(x)dx ex 2arctan ex In 2 In 2 = 0.6931 (1 + e2x) sec(2arctan ex dx Apply integral substitution. If g' is continuous on [a,b] and f is continuous on an interval Let u = g(x) and F be the antiderivative of f. containing the values of g(x) for as x s b, then F (g (x) ) = dF du ax F (u) = du dudx = f(u) = f(g(x))g'(x) ['f(o()'(x ) dx = Thus, by the fundamental theorem of Calculus, g(b) flu)du ['flow)s'(x)dx = F(9(x)12 = F(9(b)) - F(g(a)) = f(u)du g (a) Apply the exponent rule of natural logarithms. nm = eminn u = 2arctan ex dy (2arctan ex ) d dy (earctan e* In 2) Apply the chain rule: F' (x) = f'(g(x))g'(x). Let f(x) = ex, and g(x) = arctan e* In 2. As e*is its own derivative: d dx (earctan e* In 2) = (earctane*In2) _d "dx ( arctan e* In 2) The derivative of a constant multiple is equal to the multiple of a derivative: Apply integral substitution to the problem. dy ( arctan e* In 2) = In 2 -(arctan ex) Let u = garciame and solve for Apply the chain rule: F' (x) = f'(g(x))g' (x). du. Let f (x) = arctan x and g(x) = ex. d 1 d arctan x = 1+x drex =ex ex ( arctan ed) = 1 + (ex)2 d In 2 ex dy (arctan e) = 1 + e2x (earctan e* In2) = (earctane* In2) (In 2 ex) 1 + e2x Simplify. du = d 7. (2octane ) = arctan ex In 2 ( In 2 ex In 2 (2arctaner) (ex) 1 + e2x 1 + e2x -. dx