Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 38

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Evaluate $\displaystyle \int \sin (\ln x) dx$ by making a substitution first, then by using Integration by parts.

If we use $z = \ln x$, then $e^z = x$ so $dx = e^z dz$


$\displaystyle \int \sin (\ln x) dx = \int \sin z \left( e^z dz \right) = \int e^z \sin z dz$

By using Integration by parts, if we let $u = e^z$ and $dv = \sin z dz$ then,

$du = e^z dz$ and $v = -\cos z$


$ \begin{equation} \begin{aligned} \int e^z \sin z dz = uv - \int vdu &= -e^z \cos z - \int (-\cos z) (e^z dz)\\ \\ &= -e^z \cos z + \int e^z \cos z dz \end{aligned} \end{equation} $

Again by using Integration by parts, if we let

$ \begin{equation} \begin{aligned} u_1 &= e^z && \text{and}& dv_1 &= \cos z dz \text{, then}\\ \\ du_1 &= e^z dz && \text{and}& v_1 &= \sin z \end{aligned} \end{equation} $


$\displaystyle \int e^z \cos z dz = u_1 v_1 - \int v_1 du, = e^z \sin z - \int \sin z(e^z dz)$

Going back to the first equation,

$\displaystyle \int e^z \sin z dz = -e^z \cos z + \left[e^z\sin z - \int \sin z \left( e^z dz \right) \right]$

Combining like terms, we obtain

$ \begin{equation} \begin{aligned} 2 \int e^z \sin z dz &= -e^z \cos z + e^z \sin z\\ \\ \int e^z \sin z dz &= \frac{e^z(\sin z - \cos z)}{2} \end{aligned} \end{equation} $

but $z = \ln x$,


$ \begin{equation} \begin{aligned} \frac{e^z(\sin z - \cos z)}{2} &= \frac{e^{\ln x} (\sin (\ln x) - \cos (\ln x))}{2}\\ \\ &= \frac{x}{2} [ \sin (\ln x) - \cos (\ln x)] + c \end{aligned} \end{equation} $