Primitive of x over 1 minus Sine of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int \frac {x \ \mathrm d x} {1 - \sin a x} = \frac x a \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 2 {a^2} \ln \left\vert{\sin \left({\frac \pi 4 - \frac {a x} 2}\right)}\right\vert + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle \frac 1 {1 - \sin a x}\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac 1 a \tan \left({\frac \pi 4 + \frac {a x} 2}\right)\) Primitive of Reciprocal of 1 minus Sine of a x


Then:

\(\displaystyle \int \frac {x \ \mathrm d x} {1 - \sin a x}\) \(=\) \(\displaystyle x \left({\frac 1 a \tan \left({\frac \pi 4 + \frac {a x} 2}\right)}\right) - \int \left({\frac 1 a \tan \left({\frac \pi 4 + \frac {a x} 2}\right)}\right) \times 1 \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac x a \tan \left({\frac \pi 4 + \frac {a x} 2}\right) - \frac 1 a \int \tan \left({\frac \pi 4 + \frac {a x} 2}\right) \ \mathrm d x + C\) simplifying


Then:

\(\displaystyle z\) \(=\) \(\displaystyle \frac \pi 4 + \frac {a x} 2\)
\(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle \frac a 2\) Derivative of Power
\(\displaystyle \implies \ \ \) \(\displaystyle \frac 1 a \int \tan \left({\frac \pi 4 + \frac {a x} 2}\right) \ \mathrm d x\) \(=\) \(\displaystyle \frac 1 a \int \frac 2 a \tan z \ \mathrm d z\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 {a^2} \ln \left\vert{\cos z}\right\vert + C\) Primitive of Tangent Function
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 {a^2} \ln \left\vert{\cos \left({\frac \pi 4 + \frac {a x} 2}\right)}\right\vert + C\) substituting back for $z$
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 {a^2} \ln \left\vert{\sin \left({\frac \pi 2 - \left({\frac \pi 4 + \frac {a x} 2}\right)}\right)}\right\vert + C\) Sine of Complement equals Cosine
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 {a^2} \ln \left\vert{\sin \left({\frac \pi 4 - \frac {a x} 2}\right)}\right\vert + C\)


Putting it all together:

$\displaystyle \int \frac {x \ \mathrm d x} {1 - \sin a x} = \frac x a \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 2 {a^2} \ln \left\vert{\sin \left({\frac \pi 4 - \frac {a x} 2}\right)}\right\vert + C$

$\blacksquare$


Also see


Sources