Primitive of Reciprocal of Sine of a x by 1 plus Cosine of a x

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Theorem

$\ds \int \frac {\d x} {\sin a x \paren {1 + \cos a x} } = \frac 1 {2 a \paren {1 + \cos a x} } + \frac 1 {2 a} \ln \size {\tan \frac {a x} 2} + C$


Proof

Let:

\(\ds u\) \(=\) \(\ds \cos a x\)
\(\ds \frac {\d u} {\d x}\) \(=\) \(\ds -a \sin a x\) Derivative of $\cos a x$


Then:

\(\ds \int \frac {\d x} {\sin a x \paren {1 + \cos a x} }\) \(=\) \(\ds \int \frac {\sin a x \rd x} {\sin^2 a x \paren {1 + \cos a x} }\) multiplying top and bottom by $\sin a x$
\(\ds \) \(=\) \(\ds \int \frac {\sin a x \rd x} {\paren {1 - \cos^2 a x} \paren {1 + \cos a x} }\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \frac {-1} a \int \frac {\d u} {\paren {1 - u^2} \paren {1 + u} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {-1} a \int \frac {\d u} {\paren {1 - u} \paren {1 + u}^2}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \frac {-1} a \paren {\frac 1 {-2} \paren {\frac 1 {1 + u} + \frac {-1} {-2} \ln \size {\frac {1 - u} {1 + u} } } } + C\) Primitive of $\dfrac 1 {\paren {a x + b}^2 \paren {p x + q} }$
\(\ds \) \(=\) \(\ds \frac 1 {2 a \paren {1 + u} } + \frac 1 {4 a} \ln \size {\frac {1 - u} {1 + u} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {2 a \paren {1 + \cos a x} } + \frac 1 {4 a} \ln \size {\frac {1 - \cos a x} {1 + \cos a x} } + C\) substituting for $u$
\(\ds \) \(=\) \(\ds \frac 1 {2 a \paren {1 + \cos a x} } + \frac 1 {4 a} \ln \size {\frac {\frac 1 2 \sec^2 \frac {a x} 2} {\frac 1 2 \csc^2 \frac {a x} 2} } + C\) Reciprocal of $1 - \cos$ and Reciprocal of $1 + \cos$
\(\ds \) \(=\) \(\ds \frac 1 {2 a \paren {1 + \cos a x} } + \frac 1 {4 a} \ln \size {\frac {\sin^2 \frac {a x} 2} {\cos^2 \frac {a x} 2} } + C\) Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds \frac 1 {2 a \paren {1 + \cos a x} } + \frac 1 {4 a} \ln \size {\tan^2 \frac {a x} 2} + C\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac 1 {2 a \paren {1 + \cos a x} } + \frac 1 {2 a} \ln \size {\tan \frac {a x} 2} + C\) Logarithm of Power

$\blacksquare$


Also see


Sources