Primitive of Reciprocal of Power of Sine of a x

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Theorem

$\ds \int \frac {\d x} {\sin^n a x} = \frac {-\cos a x} {a \paren {n - 1} \sin^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\sin^{n - 2} a x}$


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \frac 1 {\sin^{n - 2} a x}\)
\(\ds \) \(=\) \(\ds \sin^{- n + 2} a x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds a \paren {-n + 2} \sin^{-n + 1} a x \cos a x\) Derivative of $\sin a x$, Derivative of Power, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {-a \paren {n - 2} \cos a x} {\sin^{n - 1} a x}\) simplifying


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {\sin^2 a x}\)
\(\ds \) \(=\) \(\ds \csc^2 a x\) Cosecant is $\dfrac 1 \sin$
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {-\cot a x} a\) Primitive of $\csc^2 a x$


Then:

\(\ds \int \frac {\d x} {\sin^n a x}\) \(=\) \(\ds \int \frac {\d x} {\sin^{n - 2} a x \sin^2 a x}\)
\(\ds \) \(=\) \(\ds \paren {\frac 1 {\sin^{n - 2} a x} } \paren {\frac {-\cot a x} a} - \int \paren {\frac {-\cot a x} a} \paren {\frac {-a \paren {n - 2} \cos a x} {\sin^{n - 1} a x} } \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {-\cos a x} {a \sin^{n - 1} a x} - \int \frac {\paren {n - 2} \cos^2 a x} {\sin^n a x} \rd x\) Cotangent is $\dfrac {\cos} {\sin}$
\(\ds \) \(=\) \(\ds \frac {-\cos a x} {a \sin^{n - 1} a x} - \int \frac {\paren {n - 2} \paren {1 - \sin^2 a x} } {\sin^n a x} \rd x\) Sum of $\sin^2$ and $\cos^2$
\(\ds \) \(=\) \(\ds \frac {-\cos a x} {a \sin^{n - 1} a x} - \paren {n - 2} \int \frac {\d x} {\sin^n a x} + \paren {n - 2} \int \frac {\d x} {\sin^{n - 2} a x}\) Linear Combination of Primitives
\(\ds \leadsto \ \ \) \(\ds \paren {n - 1} \int \frac {\d x} {\sin^n a x}\) \(=\) \(\ds \frac {-\cos a x} {a \sin^{n - 1} a x} + \paren {n - 2} \int \frac {\d x} {\sin^{n - 2} a x}\) gathering terms
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\sin^n a x}\) \(=\) \(\ds \frac {-\cos a x} {a \paren {n - 1} \sin^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\sin^{n - 2} a x}\) dividing by $n - 1$

$\blacksquare$


Also see


Sources

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