Primitive of Power of x by Arccosine of a x
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Theorem
- $\ds \int x^n \arccos a x \rd x = \frac {x^{n + 1} } {n + 1} \arccos a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} }$
for $n \ne -1$.
Proof
Recall:
\(\text {(1)}: \quad\) | \(\ds \int x^n \arccos x \rd x\) | \(=\) | \(\ds \frac {x^{n + 1} } {n + 1} \arccos x + \frac 1 {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - x^2} }\) | Primitive of $x^n \arccos x$ |
Then:
\(\ds \int x^n \arccos a x \rd x\) | \(=\) | \(\ds \int \dfrac 1 {a^n} \paren {a x}^n \arccos a x \rd x\) | manipulating into appropriate form | |||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {a^n} \int \paren {a x}^n \arccos a x \rd x\) | Primitive of Constant Multiple of Function | |||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {a^n} \paren {\dfrac 1 a \paren {\frac 1 {n + 1} \paren {a x}^{n + 1} \arccos a x + \frac 1 {n + 1} \int \paren {a x}^{n + 1} \frac {\d x} {\sqrt {1 - \paren {a x}^2} } } }\) | Primitive of Function of Constant Multiple, from $(1)$ | |||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{n + 1} } {n + 1} \arccos a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} }\) | simplifying
|
$\blacksquare$
Also see
Sources
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Back endpapers: A Brief Table of Integrals: $100$.