Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 20

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Determine the integral $\displaystyle \int \cos^2 x \sin 2x dx$

$ \begin{equation} \begin{aligned} \int \cos^2 x \sin 2x dx =& \int \cos^2 x (2 \sin x \cos x) dx \qquad \text{Apply Trigonometric Identity } \sin 2x = 2 \sin x \cos x \\ \\ \int \cos^2 x \sin 2x dx =& 2 \int \cos^3 x \sin x dx \end{aligned} \end{equation} $

Let $u = \cos x$, then $du = - \sin x dx$, so $\sin x dx = -du$. Thus,

$ \begin{equation} \begin{aligned} 2 \int \cos^3 x \sin x dx =& 2 \int u^3 \cdot -du \\ \\ 2 \int \cos^3 x \sin x dx =& -2 \int u^3 du \\ \\ 2 \int \cos^3 x \sin x dx =& -2 \left( \frac{u^{3 + 1}}{3 + 1} \right) + c \\ \\ 2 \int \cos^3 x \sin x dx =& \frac{-2u^4}{4} + c \\ \\ 2 \int \cos^3 x \sin x dx =& \frac{-u^4}{2} + c \\ \\ 2 \int \cos^3 x \sin x dx =& \frac{-(\cos x)^4}{2} + c \\ \\ 2 \int \cos^3 x \sin x dx =& \frac{-\cos^4 x}{2} + c \end{aligned} \end{equation} $