Integral of \( \sin^3 x \)
\( \) \( \)\( \)\( \)\( \)\( \)\( \)
Evaluate the integral
\[ \int \sin^3 x \; dx \]
Write the integrand \( \sin^3 x \) as a product of as \( \sin^2 x \) and \( \sin x \)
\[ \int \sin^3 x \; dx = \int \sin^2 x \sin x \; dx\]
Use of the trigonometric identity \( \; \sin^2 x = 1 - \cos^2 x \) to write the integral as
\[ \int \sin^3 x \; dx = \int (1 - \cos^2 x) \sin x \; dx\]
Expand \( (1 - \cos^2 x) \sin x \) and write
\[ \int \sin^3 x \; dx = \int \sin x \; dx - \int \cos^2 x \sin x \; dx \]
Integration by Substitution : Let \( u = \cos x \) and hence \( \dfrac{du}{dx} = - \sin x \) or \( du = - \sin x \; dx \) and substitute to obtain
\[ \int \sin^3 x \; dx = \int \sin x \; dx + \int u^2 \; du \]
Use integral formulas to evaluate the above integral and write
\[ \int \sin^3 x \; dx = - \cos x + \dfrac{1}{3} u^3 + c \]
Substitute back \( u = \cos x \) to find the final answer
\[ \boxed { \int \sin^3 x \; dx = - \cos x + \dfrac{1}{3} \cos^3 x + c } \]
More References and Links
- Table of Integral Formulas
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University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
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Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
- Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8