# Recall that \sin ^{-1}x=\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}dt Find the first four nonzero terms...

## Question:

Recall that {eq}\sin ^{-1}x=\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}dt {/eq} Find the first four nonzero terms in the Maclaurin series for {eq}sin^{-1} x. {/eq}

## Power Series; Binomial theorem:

{eq}\\ {/eq}

To get the power series representation of the inverse sine function, we will use the derivative property that itself is mentioned in the problem. First of all, we will use the Binomial series for the fractional negative powers in order to determine the power series for the integrand function.

{eq}\displaystyle (1 + a)^{n} = 1 + na + \dfrac {n(n -1)}{2!} \; a^{2} + \dfrac {n (n -1) (n -2)}{3!} \; a^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |a| < 1 {/eq}

{eq}\displaystyle \int_{}^{} \; t^{n} \; dx = \biggr( \dfrac {t^{n+1}}{n+1} \biggr) + \text {C} {/eq}

{eq}\text {C is the constant of integration} {/eq}

## Answer and Explanation:

{eq}\\ {/eq}

{eq}\displaystyle (1 + a)^{-\dfrac {1}{2}} = 1 - \dfrac {a}{2} + \dfrac {\biggr( -\dfrac {1}{2} \biggr) \; \biggr(- \dfrac {1}{2} - 1 \biggr)}{2!} \; a^{2} + \dfrac {\biggr(- \dfrac {1}{2} \biggr) \; \biggr(- \dfrac {1}{2} - 1 \biggr) \; \biggr(- \dfrac {1}{2} - 2 \biggr)}{3!} \; a^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow |a| < 1 {/eq}

{eq}\displaystyle \dfrac {1}{\sqrt {1 + a}} = 1 - \dfrac {a}{2} + \dfrac {3a^{2}}{8} - \dfrac {5a^{3}}{16} + \cdots {/eq}

Now replace the value of {eq}\; a = - t^{2} \; {/eq} in the above expression:

{eq}\displaystyle \dfrac {1}{\sqrt {1 - t^{2}}} = 1 + \dfrac {t^{2}}{2} + \dfrac {3t^{4}}{8} + \dfrac {5t^{6}}{16} + \cdots {/eq}

{eq}\displaystyle \int_{0}^{x} \; \dfrac {dt}{\sqrt {1 - t^{2}}} = \int_{0}^{x} \; \; \Biggr[ 1 + \dfrac {t^{2}}{2} + \dfrac {3t^{4}}{8} + \dfrac {5t^{6}}{16} + \cdots \Biggr] \; dt {/eq}

{eq}\displaystyle \arcsin(x) = \Biggr[ t + \dfrac {t^{3}}{3 \times 2} + \dfrac {3t^{5}}{5 \times 8} + \dfrac {5t^{7}}{7 \times 16} + \cdots \Biggr]_{t = 0}^{t = x} {/eq}

Finally, the power series representation of the function {eq}\; \arcsin(x) \; {/eq} is given as:

{eq}\displaystyle \Longrightarrow \boxed {\arcsin(x) = x + \dfrac {x^{3}}{6} + \dfrac {3x^{5}}{40} + \dfrac {5x^{7}}{112} + \cdots} {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow |x| < 1 {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from Algebra II Textbook

Chapter 21 / Lesson 16