antiderivative of sec^2x
The antiderivative of the function f(x) = sec^2(x) can be found by using the reverse of the power rule for derivatives
The antiderivative of the function f(x) = sec^2(x) can be found by using the reverse of the power rule for derivatives. The reverse of the power rule states that if F'(x) = f(x), then the antiderivative F(x) of f(x) is given by:
F(x) = ∫f(x) dx
To find the antiderivative of sec^2(x), let’s rewrite it in terms of trigonometric identities. We know that sec^2(x) = 1/cos^2(x). Using the identity cos^2(x) = 1 – sin^2(x), we can rewrite sec^2(x) as:
sec^2(x) = 1/(1 – sin^2(x))
Now, we can perform a trigonometric substitution. Let’s substitute sin(x) with u, so du = cos(x) dx:
F(x) = ∫(1/(1 – sin^2(x))) dx
Let u = sin(x), then du = cos(x) dx
F(x) = ∫(1/(1 – u^2)) du
We can rewrite the integrand by using a partial fraction decomposition:
1/(1 – u^2) = A/(1 – u) + B/(1 + u)
To solve for A and B, we need to find a common denominator:
1/(1 – u^2) = (A(1 + u) + B(1 – u))/((1 – u)(1 + u))
1 = (A(1 + u) + B(1 – u))
We expand this equation and match the coefficients of the powers of u:
1 = (A + B) + (A – B)u
Comparing the coefficients, we find that A + B = 1 and A – B = 0. Solving these equations, we get A = B = 1/2.
Now, we can rewrite the integrand again:
1/(1 – u^2) = (1/2)(1/(1 – u) + 1/(1 + u))
Using this partial fraction decomposition, we can rewrite the antiderivative as:
F(x) = ∫[(1/2)(1/(1 – u) + 1/(1 + u))] du
F(x) = (1/2) ∫[(1/(1 – u) + 1/(1 + u))] du
To solve the integral, we can break it down into two separate integrals:
F(x) = (1/2) ∫[1/(1 – u)] du + (1/2) ∫[1/(1 + u)] du
We can solve these two separate integrals using the natural logarithm function:
F(x) = (1/2) ln|1 – u| + (1/2) ln|1 + u| + C
Next, we substitute u back in for sin(x):
F(x) = (1/2) ln|1 – sin(x)| + (1/2) ln|1 + sin(x)| + C
Therefore, the antiderivative of sec^2(x) is:
F(x) = (1/2) ln|1 – sin(x)| + (1/2) ln|1 + sin(x)| + C
where C is the constant of integration.
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