[amp_mcq option1=”0″ option2=”1″ option3=”$$l$$n 2″ option4=”$$\\frac{1}{2}l$$n 2″ correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\frac{\\pi}{4}}$.<\/p>\n
We can evaluate the integral by substituting $u = \\tan x$, which gives $du = \\sec^2 x \\, dx$. Substituting gives us:<\/p>\n
$$\\int_0^{\\frac{\\pi}{4}} \\frac{(1 – \\tan x)}{(1 + \\tan x)} \\, dx = \\int_0^1 \\frac{(1 – u)}{(1 + u)} \\sec^2 x \\, du$$<\/p>\n
We can now evaluate this integral using partial fractions:<\/p>\n
$$\\int_0^1 \\frac{(1 – u)}{(1 + u)} \\sec^2 x \\, du = \\int_0^1 \\left( \\frac{1}{2} – \\frac{1}{1 + u} \\right) \\sec^2 x \\, du$$<\/p>\n
The first integral is easy to evaluate:<\/p>\n
$$\\int_0^1 \\frac{1}{2} \\sec^2 x \\, du = \\frac{\\sec^2 x}{2} \\bigg|_0^1 = \\frac{1}{2}$$<\/p>\n
The second integral can be evaluated using the substitution $v = 1 + u$, which gives $du = -dv$. Substituting gives us:<\/p>\n
$$\\int_0^1 \\frac{-1}{v} \\sec^2 x \\, du = -\\int_1^2 \\frac{1}{v} \\sec^2 x \\, dv$$<\/p>\n
We can now evaluate this integral using the following identity:<\/p>\n
$$\\int \\frac{\\sec^2 x}{v} \\, dv = \\ln |v| + \\tan x + C$$<\/p>\n
where $C$ is an arbitrary constant. Substituting gives us:<\/p>\n
$$\\int_0^1 \\frac{-1}{v} \\sec^2 x \\, dv = -\\ln |v| – \\tan x \\bigg|_1^2 = -\\ln 2 – \\tan x + \\ln 2 + \\tan x = -\\tan x$$<\/p>\n
Therefore, the integral we are looking for is:<\/p>\n
$$\\int_0^{\\frac{\\pi}{4}} \\frac{(1 – \\tan x)}{(1 + \\tan x)} \\, dx = \\frac{1}{2} – \\left( -\\tan x \\right) = \\frac{\\pi}{4}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0″ option2=”1″ option3=”$$l$$n 2″ option4=”$$\\frac{1}{2}l$$n 2″ correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20228","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n