Primitive of x over 1 plus Sine of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int \frac {x \rd x} {1 + \sin a x} = -\frac x a \map \tan {\frac \pi 4 - \frac {a x} 2} + \frac 2 {a^2} \ln \size {\map \sin {\frac \pi 4 + \frac {a x} 2} } + C$


Proof

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac 1 {1 + \sin a x}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle -\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2}\) Primitive of $\dfrac 1 {1 + \sin a x}$


Then:

\(\displaystyle \int \frac {x \rd x} {1 + \sin a x}\) \(=\) \(\displaystyle x \paren {-\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2} } - \int \paren {-\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2} } \times 1 \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle -\frac x a \map \tan {\frac \pi 4 - \frac {a x} 2} + \frac 1 a \int \map \tan {\frac \pi 4 - \frac {a x} 2} \rd x + C\) simplifying


Then:

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


Putting it all together:

$\displaystyle \int \frac {x \rd x} {1 - \sin a x} = -\frac x a \map \tan {\frac \pi 4 - \frac {a x} 2} + \frac 2 {a^2} \ln \size {\map \sin {\frac \pi 4 + \frac {a x} 2} } + C$

$\blacksquare$


Also see


Sources