Integral of tanx

The integral of tangent function cannot be expressed in terms of elementary functions. However, it can be expressed in terms of natural logarithmic functions by using the following substitution:

Let u = cos(x), then du/dx = -sin(x) and

So, sin(x)dx = – du

∫ tan(x) dx

= ∫ [(sin(x)/cos(x))]dx

= ∫ [(-1/u) du]

= – ln|u|

= – ln|cos(x)| + C (Ans.)

= ln |cos(x)|⁻¹ + C

= ln |1/cos(x)| + C

= ln |sec(x)| + C (Ans.)

where C is the constant of integration.

Therefore, the indefinite integral of tan(x) is – ln|cos(x)| + C.

Integral of tan(nx)

To integrate tan(nx), we can use the substitution:

u = nx

du/dx = n

dx = du/n

Substituting u and dx in the integral, we get:

∫ tan(nx) dx = (1/n) ∫ tan(u) du

Now, we can use the identity:

tan(x) = sin(x)/cos(x)

to write:

∫ tan(nx) dx = (1/n) ∫ sin(u)/cos(u) du

Using the substitution:

v = cos(u),

dv = -sin(u) du,

we get:

∫ tan(nx) dx = (-1/n) ∫ (1/v) dv

∫ tan(nx) dx = (-1/n) ln|v| + C

Substituting back v = cos(u) and u = nx, we get:

∫ tan(nx) dx = (-1/n) ln|cos(nx)| + C

Therefore, the integral of tan(nx) is (-1/n) ln|cos(nx)| + C

Alternative Method

The integral of tan(nx) can be found by using the substitution method.

Let u = nx, then du/dx = n and dx = du/n. Substituting these into the integral, we get:

∫ tan(nx) dx = (1/n) ∫ tan(u) du

We can now use the formula for the integral of tan(x):

∫ tan(x) dx = ln|sec(x)| + C

where C is the constant of integration.

Substituting back, we have:

∫ tan(nx) dx = (1/n) ∫ tan(u) du = (1/n) ln|sec(u)| + C

Substituting u = nx, we get:

∫ tan(nx) dx = (1/n) ln|sec(nx)| + C

Therefore, the integral of tan(nx) is (1/n) ln|sec(nx)| + C, where C is the constant of integration.