integral of (tanx)^4
(tanx)^4 = (tanx)^2 (tanx)^2
=(sec^2 x - 1)(tan^2 x)
=(sec^2 x)(tan^2 x) - tan^2 x
= integral of sec^2 x tan^2 x dx - integral of tan^2 x dx
First, integral of sec^2 x tan^2 x dx
Let u = tanx because that would make du = sec^2 x dx
so then we have integral of u^2 du which is (1/3)u^3
substituting back in tanx we get (1/3)tan^3 x
Next, integral of tan^2 x
tan^2 x = sec^2 x -1
integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx
= tanx - x
so putting it all together we have integral of tan^4 x dx =
(1/3)tan^3 x - tanx + x + C
aln(absolute value secax) + C
It isn't quite clear what you mean with "underoot".
The integral of sec(x) with respect to x is ln|sec(x) + tan(x)| + C, where C is the constant of integration. This result can be derived using integration techniques such as substitution or integration by parts. The integral of sec(x) is a common integral in calculus and is often used in trigonometric integrals.
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
∫ tan(x) dx = -ln(cos(x)) + C C is the constant of integration.
I googled that and found PDF's of the electrical wiring diagrams for my Corolla. The integration relay is located to the left of the steering column behind the coin tray attached to the fuse box. It should be tan in color.
Will try integration by parts. uv - int[v du] u = sec(x)----------------du = sec(x) tan(x) dv = tan(x)---------------v = ln[sec(x)] sec(x) ln[sex(x)] - int[lnsec(x) dx] = sec(x) ln[sec(x)] - xlnsec(x) - x + C ===========================
The integral of sqrt(tan(x)) is rather complex and is hard to show with the formatting allowed on Answers.com. See the related links for a representation of the answer.
∫ 1/cos2(x) dx = tan(x) + C C is the constant of integration.
∫ 1/sin(x) dx = ln(tan(x/2)) + C C is the constant of integration.
∫ 1/cos(x) dx = ln(sec(x) + tan(x)) + C C is the constant of integration.
INTEGRATION