How do you integrate #int sec^-1x# by integration by parts method?
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The answer is
We need
Integration by parts is
Here, we have
Therefore,
Perform the second integral by substitution
So,
Finally,
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To integrate ∫sec^(-1)(x) dx using integration by parts method, you can set up the integral as follows:
Let u = sec^(-1)(x) and dv = dx. Then, differentiate u to find du and integrate dv to find v.
Differentiating u: du = (1 / (x√(x^2 - 1))) dx
Integrating dv: v = x
Now, apply the integration by parts formula:
∫u dv = uv - ∫v du
Substitute the values of u, dv, du, and v into the formula:
∫sec^(-1)(x) dx = x * sec^(-1)(x) - ∫x * (1 / (x√(x^2 - 1))) dx
Now, integrate the remaining integral, which is relatively simpler:
∫x * (1 / (x√(x^2 - 1))) dx
Letting u = x^2 - 1 and du = 2x dx, we have:
(1/2) * ∫(1 / √u) du
This integrates to:
(1/2) * 2 * √u + C
= √(x^2 - 1) + C
Substitute this back into the original equation:
∫sec^(-1)(x) dx = x * sec^(-1)(x) - √(x^2 - 1) + C
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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