Derivation and Evaluation
Evaluate the integral:
\[ \int \sin^2(x) \cos^3(x) \, dx \]Rewrite the integral by splitting off a single cosine factor:
\[ = \int \sin^2(x) \cos^2(x) \cos x \, dx \]Use the trigonometric identity \( \cos^2 x = 1 - \sin^2 x \) and substitute:
\[ = \int \sin^2(x) (1 - \sin^2(x)) \cos x \, dx \]Expand the integrand:
\[ = \int (\sin^2(x) - \sin^4(x)) \cos x \, dx \]Use Integration by Substitution: let \( u = \sin x \), which gives \( du = \cos x \, dx \). Substituting this yields:
\[ = \int (u^2 - u^4) \, du \]Use the common integral formula \( \int u^n \, du = \dfrac{1}{n+1} u^{n+1} + c \) to evaluate the integral:
\[ \dfrac{1}{3} u^3 - \dfrac{1}{5} u^5 + c \]where \( c \) is the constant of integration.
Substitute back \( u = \sin x \) to obtain the final answer:
Integral Formula for \( \sin^2(x)\cos^3(x) \):
\[ \int \sin^2(x) \cos^3(x) \, dx = \dfrac{1}{3} \sin^3 x - \dfrac{1}{5} \sin^5 x + c \]
More References and Links
- Table of Integral Formulas
- University Calculus - Early Transcendentals - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13: 978-0134995540
- Calculus - Gilbert Strang - MIT - ISBN-13: 978-0961408824
- Calculus - Early Transcendentals - James Stewart - ISBN-13: 978-0-495-01166-8