Primitive of Reciprocal of Square of 1 plus Sine of a x

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Theorem

$\ds \int \frac {\d x} {\paren {1 + \sin a x}^2} = \frac {-1} {2 a} \map \tan {\frac \pi 4 - \frac {a x} 2} - \frac 1 {6 a} \map {\tan^3} {\frac \pi 4 - \frac {a x} 2} + C$


Proof

\(\ds \int \frac {\d x} {\paren {1 + \sin a x}^2}\) \(=\) \(\ds \int \paren {\frac 1 2 \map {\sec^2} {\dfrac \pi 4 - \frac {a x} 2} }^2 \rd x\) Reciprocal of One Plus Sine
\(\ds \) \(=\) \(\ds \frac 1 4 \int \map {\sec^4} {\frac \pi 4 - \frac {a x} 2} \rd x\) simplifying

Let:

\(\ds z\) \(=\) \(\ds \frac \pi 4 - \frac {a x} 2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\ d x}\) \(=\) \(\ds \frac {-a} 2\) simplifying


Thus:

\(\ds \frac 1 4 \int \map {\sec^4} {\frac \pi 4 - \frac {a x} 2} \rd x\) \(=\) \(\ds \frac 1 4 \int \frac {-2} a \sec^4 z \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {-1} {2 a} \int \sec^4 z \rd z\) simplifying
\(\ds \) \(=\) \(\ds \frac {-1} {2 a} \paren {\frac {\sec^2 z \tan z} 3 + \frac 2 3 \int \sec^2 z \rd z} + C\) Primitive of $\sec^n a x$
\(\ds \) \(=\) \(\ds \frac {-1} {2 a} \paren {\frac {\sec^2 z \tan z} 3 + \frac 2 3 \tan z} + C\) Primitive of $\sec^2 a x$
\(\ds \) \(=\) \(\ds \frac {-1} {2 a} \paren {\frac {\paren {1 + \tan^2 z} \tan z} 3 + \frac 2 3 \tan z} + C\) Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \frac {-1} {2 a} \tan z - \frac 1 {6 a} \tan^3 z + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {-1} {2 a} \map \tan {\frac \pi 4 - \frac {a x} 2} - \frac 1 {6 a} \map {\tan^3} {\frac \pi 4 - \frac {a x} 2} + C\) substituting for $z$

$\blacksquare$


Also see


Sources