Integral of \( \sin^2(x) \cos^2(x) \)

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Evaluate the integral \[ \int \sin^2(x) \cos^2(x) \; dx \]
Use
trigonometric identities \( \sin^2 x = \dfrac{1}{2}(1 - \cos (2 x)) \) and \( \cos^2 x = \dfrac{1}{2}(1 + \cos (2 x)) \) on the right and write
\[ \int \sin^2(x) \cos^2(x) \; dx = \dfrac{1}{4} \int (1 - \cos (2 x)) (1 + \cos (2 x)) \; dx \]
Expand the integrand of the integral on the right side \[ = \dfrac{1}{4} \int (1 - \cos^2(2 x)) \; dx \] Use
trigonometric identity \( \cos^2 x = \dfrac{1}{2}(1 + \cos (2 x)) \) which also gives \( \cos^2 (2x) = \dfrac{1}{2}(1 + \cos (4x)) \)
\[ = \dfrac{1}{8} \int (1 - \dfrac{1}{2}(1 + \cos (4x)) ) \; dx \] Expand and simplify the integrand \[ = \dfrac{1}{8} \int (1 - \dfrac{1}{2}(1 - \cos (4x)) ) \; dx \] Use the common integrals \( \int 1 dx = x + c\) and \( \int \cos (kx) dx = \dfrac{1}{k} \sin x + c\) to evaluate the above integral and obtain the final answer \[ \boxed { \int \sin^2(x) \cos^2(x) \; dx = \dfrac{1}{8} \left(x - \dfrac{1}{4} \sin (4x) \right) + c } \]



More References and Links

  1. Table of Integral Formulas
  2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
  3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
  4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8