∫sec²xdx
To integrate ∫sec²xdx, we can use the power rule of integration
To integrate ∫sec²xdx, we can use the power rule of integration. The power rule states that if we have an expression of the form ∫x^n dx, where n is any real number except -1, the integral evaluates to (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
In the case of ∫sec²xdx, we can rewrite it as the integral of (cos²x)^(-1) dx. Now, let’s consider the integral as follows:
∫(cos²x)^(-1) dx
To simplify this expression, we can use the identity: cos²x = 1 – sin²x.
This gives us:
∫(1 – sin²x)^(-1) dx
Now, let’s use the substitution method. Let u = sin(x), which implies du = cos(x) dx. We can rewrite the integral as:
∫(1 – u²)^(-1) du
Now, we have an expression of the form ∫(1 – u²)^(-1) du, which we can integrate using partial fractions. We can break it down as:
1/(1 – u²) = A/(1 – u) + B/(1 + u)
To find the values of A and B, we can multiply both sides by (1 – u)(1 + u):
1 = A(1 + u) + B(1 – u)
Expanding this equation, we get:
1 = A + Au + B – Bu
Equating the coefficients of u and the constant terms, we have:
1 = A + B [coefficient of u]
0 = A – B [constant terms]
From the second equation, we can solve for A in terms of B:
A = B
Substituting this back into the first equation, we have:
1 = 2A
Therefore, A = 1/2 and B = 1/2.
Now, let’s rewrite the integral using the partial fraction decomposition:
∫(1 – u²)^(-1) du = (1/2) * ∫(1/(1 – u)) du + (1/2) * ∫(1/(1 + u)) du
Evaluating each integral separately, we get:
(1/2) * ln|(1 – u)| + (1/2) * ln|(1 + u)| + C
Substituting back u = sin(x), we get:
(1/2) * ln|(1 – sin(x))| + (1/2) * ln|(1 + sin(x))| + C
Therefore, the integral of ∫sec²xdx is:
(1/2) * ln|(1 – sin(x))| + (1/2) * ln|(1 + sin(x))| + C, where C is the constant of integration.
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