What is the integral of ${sec}^{- 1} (x)$ $sec^-1(x)$ ?

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Answer 1 · 2014-12-18T01:31:02

We apply the reciprocal properties of trigonometric functions.

By definition,

${sec}^{- 1} x = \frac{1}{sec x}$ $sec^-1 x = 1/sec x$

Therefore, the $\int {sec}^{- 1} x d x$ $int sec^-1 x dx$ is

$\int (\frac{1}{sec x}) d x$ $int (1/sec x) dx$

We know that $\frac{1}{sec x} = cos x$ $1/sec x = cos x$

Thus,

$\int cos x d x$ $int cos x dx$

$sin x + c$ $sin x + c$

What is the integral of sec−1(x)sec^-1(x)?