Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 10

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Evaluate $\displaystyle \int \sin^{-1} x dx$

If we let $u = \sin^{-1} x$ and $dv = dx$, then

$\displaystyle du = \frac{1}{\sqrt{1-x^2}} dx$ and $\displaystyle v = \int dx = x$

So,

$\displaystyle \int \sin^{-1} x dx = uv - \int v du = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx$

To evaluate $\displaystyle \int \frac{x}{\sqrt{1-x^2}} dx$ we let $u_1 = 1 - x^2$, then $du_1 = -2xdx$

Then,

$ \begin{equation} \begin{aligned} \int \frac{x}{\sqrt{1-x^2}} dx &= \int \frac{\frac{-du_1}{2}}{\sqrt{u_1}} = \frac{-1}{2} \int u_1^{-\frac{1}{2}} du_1 = -\frac{1}{2} \left[ \frac{u_1^{\frac{1}{2}}}{\frac{1}{2}} \right]\\ \\ &= -(u_1)^{\frac{1}{2}} + c = -(1-x^2)^{\frac{1}{2}} + c \end{aligned} \end{equation} $

Therefore,

$ \begin{equation} \begin{aligned} \int \sin^{-1} x dx &= x \sin^{-1} x - \left[ -(1-x^2)^{\frac{1}{2}} \right] + c\\ \\ &= x \sin^{-1} x + (1-x^2)^{\frac{1}{2}} + c \end{aligned} \end{equation} $