Solving the Integral of sec(x) tan(x) using Substitution Method

secx tanx dx

To integrate sec(x) tan(x) dx, we can use a substitution method

To integrate sec(x) tan(x) dx, we can use a substitution method. Let’s assume u = sec(x) + tan(x).

Differentiating both sides with respect to x, we get:
du/dx = sec(x)tan(x) + sec^2(x)

Now, let’s rearrange and solve for dx:
dx = du / (sec(x)tan(x) + sec^2(x))

Substituting this value of dx in the integral, we have:
∫sec(x) tan(x) dx = ∫(1 / (sec(x)tan(x) + sec^2(x))) du

Now, let’s simplify the integral:
∫(1 / (sec(x)tan(x) + sec^2(x))) du = ∫(1 / (u + u^2)) du

Splitting the fraction, we can write:
∫(1 / (u + u^2)) du = ∫(1/u) / (1 + u) du

To integrate this, we can use the method of partial fractions:
1/u = (A / u) + (B / (1 + u))

Multiplying through by u(1 + u), we get:
1 = A(1 + u) + Bu

Expanding and equating coefficients, we obtain:
1 = (A + B) + (Au + Bu)

Comparing coefficients, we have:
A + B = 0 (Equation 1)
A + B = 1 (Equation 2)

By solving these equations simultaneously, we find A = 1/2 and B = -1/2.

Now, let’s substitute these values back into the integral:
∫(1/u) / (1 + u) du = ∫((1/2) / u – (1/2) / (1 + u)) du

Using the linearity of integration, we can split this into two separate integrals:
∫((1/2) / u – (1/2) / (1 + u)) du = (1/2) * ln|u| – (1/2) * ln|1 + u|

Substituting back the value of u = sec(x) + tan(x), we get:
= (1/2) * ln|sec(x) + tan(x)| – (1/2) * ln|1 + sec(x) + tan(x)| + C

Thus, the final solution to the integral ∫sec(x) tan(x) dx is:
(1/2) * ln|sec(x) + tan(x)| – (1/2) * ln|1 + sec(x) + tan(x)| + C, where C is the constant of integration.

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