Primitive of Reciprocal of Square of 1 plus Sine of a x
Jump to navigation
Jump to search
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$: $14.359$