Integration bothering me?

Question

the indefinite integral of 1/[ (tanx)^2 + (secx)^2 ] dx
this has been bothering me for hours. I have tried everything I know, Trig subs, long division, subs, impartial fractions... Plz help!!! This is not a homework problem, just doing it for fun. (Hard to believe, sometimes i'd rather do this than video games.)

Answer 1

This one was fun.

∫1/(tan² x + sec² x) dx

sec² x = tan² x + 1, so:

∫1/(2 tan² x + 1) dx

Multiply both numerator and denominator by cos² x:

∫cos² x/(2 sin² x + cos² x) dx

Using the Pythagorean identity:

∫cos² x/(sin² x + 1) dx

Again with Pythagoras:

∫(1-sin² x)/(sin² x + 1) dx

Rewriting this as the sum of two fractions:

∫-(sin² x + 1)/(sin² x + 1) + 2/(sin² x + 1) dx

Simplifying:

∫-1 + 2/(sin² x + 1) dx

Dividing both numerator and denominator by cos² x:

∫-1 + 2 sec² x/(tan² x + sec² x) dx

Simplifying:

∫-1 + 2 sec² x/(2 tan² x + 1) dx

Break apart this integral:

-x + 2∫sec² x/(2 tan² x + 1) dx

Let u=√2 tan x, du=√2 sec² x dx

-x + √2∫1/(u² + 1) du

Integrate:

-x + √2 arctan (√2 tan x) + C

And we are done.

Answer 2

Integrate ∫{1/(tan²x + sec²x)} dx
Let
u = tanx
du = sec²x dx
du/sec²x = dx
u² = tan²x
u² + 1 = sec²x
du/(u² + 1) = dx

∫{1/(tan²x + sec²x)} dx
= ∫{[1/(u² + 1)] / [u² + (u² + 1)]} du
= ∫{1 / (u² + 1)(2u² + 1)} du

Partial fraction decomposition yields

= ∫{-1/(u² + 1)du + ∫2/(2u² + 1)du

First integral
tanθ = u
sec²θdθ = du

Second integral
(1/√2)tanα = u
(1/√2)sec²αdα = du

= ∫{-1/(u² + 1)du + ∫2/(2u² + 1)du
= ∫{-sec²θ/(tan²θ + 1)dθ + ∫(√2)sec²α/(tan²α + 1)dα
= ∫{(-sec²θ/sec²θ)dθ + ∫(√2)(sec²α/sec²α)dα
= -∫{dθ + ∫(√2)dα
= -θ + (√2)α + C
= -arctan(u) + (√2)arctan(u√2) + C
= -arctan[tan(x)] + (√2)arctan[(√2)tan(x)] + C
= -x + (√2)arctan[(√2)tan(x)] + C

Answer 3

i imagine the realm below the curve is in simple terms int (function) dx dx is ordinary, in the journey that your discern is sweet it in simple terms is going from -2 to 2 the realm above the curve is the realm of the rectangle - the realm below. i imagine that you do not favor any complicated length. per chance I neglected the point. :)

Answer 4

its a beast. you really, really, really don't want to do that by hand.

1/3*[((sqrt(3)*arctan(sqrt(3)* cos(x)+2*sin(x))/3*cos(x)) -sqrt(3)*arctan(sqrt(3)* (cos(x) - 2*sin(x))/3*cos(x)) - sqrt(3) modulus(2*x - Pi, 2*Pi) + (2*sqrt(3)-3)] + Pi/2