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Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 6

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Determine the integral $\displaystyle \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx$

Let $u = \sqrt{x}$, then $\displaystyle du = \frac{1}{2 \sqrt{x}} dx$, so $\displaystyle \frac{1}{\sqrt{x}} dx = 2du$. Thus,

$ \begin{equation} \begin{aligned} \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx =& int \sin^3 (\sqrt{x}) \cdot \frac{1}{\sqrt{x}} dx \\ \\ \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx =& \int \sin^3 u \cdot 2du \\ \\ \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx =& 2 \int \sin^3 u du \\ \\ \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx =& 2 \int \sin^2 u \sin u du \qquad \text{Apply Trigonometric Idendities } \sin^2 x = 1 - \cos^2 x \\ \\ \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx =& 2 \int (1 - \cos^2 u) \sin u du \end{aligned} \end{equation} $

Let $v = \cos u$, then $dv = - \sin u du$, so $\sin u du = -dv$. Thus,

$ \begin{equation} \begin{aligned} 2 \int (1 - \cos^2 u) \sin u du =& 2 \int (1 - v^2) \cdot -dv \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& - 2 \int (1 - v^2) dv \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& -2 \left( v - \frac{v^{2 + 1}}{2 + 1} \right) + c \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& -2 \left( v - \frac{v^3}{3} \right) + c \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& -2 \left( \cos u - \frac{\cos ^3 u}{3} \right) + c \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& -2 \left[ \cos (\sqrt{x}) - \frac{\cos^3 (\sqrt{x})}{3} \right] + c \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& -2 \cos (\sqrt{x}) + \frac{2 \cos^3 (\sqrt{x})}{3} + c \\ \\ \text{or} & \\ \\ 2 \int (1 - \cos^2 u) \sin u du =& \frac{2 \cos ^3 (\sqrt{x})}{3} - 2 \cos (\sqrt{x}) + c \end{aligned} \end{equation} $