How do you find the definite integral of #tan x dx# from #[pi/4, pi]#?

Question

Ezekiel Alexander · Accepted Answer

Actual Value: Impossible to determine

Value by Cauchy's Principal Value: #int_(pi/4)^pitanxdx = ln(1/sqrt(2)) = -0.34657# For a simple answer, this integral doesn't converge (it has an asymptote at #x = pi/2#. However, if we use Cauchy principal value, we can derive the above answer. We are asked to find #int_(pi/4)^pi tanx dx#. The first step in solving this problem is determining the antiderivative. Recall that #tanx = sinx/cosx#: #= int_(pi/4)^pi sinx/cosx dx# This is currently very hard to integrate. Let's attempt a u-substitution. Let #u = cosx#. Then #du = -sinxdx# and #dx = -(du)/sinx# #=int_(pi/4)^pi sinx/u * -(du)/sinx# #=int_(pi/4)^pi -1/udu# By the rule #int1/xdx = ln|x| + C#. #=-[ln|u|]_(pi/4)^pi# #= -[ln|cosx|]_(pi/4)^pi# Evaluate using the second fundamental theorem of calculus, which states that #int_a^bF(x)dx = f(b) - f(a)#, where #int(F(x))dx = f(x)#. #=-(ln|cospi| - ln|cospi/4|)# #~= -0.34657# Hopefully this helps!

Isabella Ackerman · Answer

undefined To find the definite integral of tan(x) dx from π/4 to π, you can use the properties of the tangent function and integration techniques. Firstly, rewrite tan(x) in terms of sine and cosine using the identity tan(x) = sin(x)/cos(x). Then, apply the substitution method, letting u = cos(x) and du = -sin(x) dx. This transforms the integral into ∫(1/u) du. Integrate this expression with respect to u, which yields ln|u|. Now, substitute back u = cos(x), resulting in ln|cos(x)|. Finally, evaluate this expression from π/4 to π and subtract the lower limit value from the upper limit value. This process will give you the definite integral of tan(x) dx from π/4 to π.