Inverse function integration formula
From Calculus
Contents
Statement
In terms of indefinite integrals
Suppose is a continuous one-one function. Then, we have:
where .
More explicitly, if is an antiderivative for
, then:
This can be justified either directly or using integration by parts and integration by u-substitution.
In terms of definite integrals
Suppose is a continuous one-one function on an interval. Suppose we are integrating
on an interval of the form
that lies in the range of
. Then:
This simplifies to:
More explicitly, if is an antiderivative for
, then:
Significance
Computational feasibility significance
Version type | Significance |
---|---|
indefinite integral | Given an antiderivative for a continuous one-one function ![]() ![]() ![]() ![]() |
definite integral | Given an antiderivative for a continuous one-one function ![]() ![]() ![]() ![]() ![]() ![]() |
Examples
Original function | Domain on which it restricts to a one-one function | Inverse function for the restriction to that domain | Domain of inverse function (equals range of original function) | Antiderivative of original function | Antiderivative of inverse function | Explanation using inverse function integration formula | Alternate explanation using integration by parts |
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sine function ![]() |
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arc sine function ![]() |
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negative of cosine function, i.e., ![]() |
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We get ![]() ![]() ![]() ![]() ![]() |
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tangent function ![]() |
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arc tangent function ![]() |
all real numbers | ![]() ![]() |
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We get ![]() ![]() |
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exponential function ![]() |
all real numbers | natural logarithm ![]() |
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exponential function ![]() |
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We get ![]() |
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