Question

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{u}}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{d}\mathbf{x}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$

Short answer

${\int }_0^u\frac{\sin xdx}{\sqrt{1-x^2}}=\frac{u^2}{2}+\frac{u^4}{12}+\frac{u^6}{20}+\dots$

Step-by-step solution

Given information

The Maclaurin expansion of the function${\int }_0^u\frac{\sin xdx}{\sqrt{1-x^2}}$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$\frac{1}{\sqrt{1-x^2}}=1+\frac{x^2}{2}+\frac{3x^4}{8}+\frac{5x^6}{16}+\dots$

$sin\left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

Part-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series.

$\frac{1}{\sqrt{1-x^2}}=1+\frac{x^2}{2}+\frac{3x^4}{8}+\frac{5x^6}{16}+\dots$

$\sin \left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

Part-B Step 3: Use the given information to define another Maclaurin series

Find the Maclaurin series of$\sin \left(x\right)\times \frac{1}{\sqrt{1-x^2}}$

Use the defined to series and evaluate.

$\sin \left(x\right)\times \frac{1}{\sqrt{1-x^2}}=\left[x-\frac{x^3}{6}+\frac{x^5}{120}+\dots \right]\times \left[1+\frac{x^2}{2}+\frac{3x^4}{8}+\dots \right]$

$\sin \left(x\right)\times \frac{1}{\sqrt{1-x^2}}=\frac{960x+320x^3+288x^5}{960}+\dots$

$\sin \left(x\right)\times \frac{1}{\sqrt{1-x^2}}=x+\frac{x^3}{3}+\frac{3x^5}{10}+\dots$

Part-C Step 3: Integrate the series to arrive at the solution

Integrate the series.

${\int }_0^u\frac{\sin xdx}{\sqrt{1-x^2}}={\int }_0^u\left[x+\frac{x^3}{3}+\frac{3x^5}{10}+\dots \right]dx$

${\int }_0^u\frac{\sin xdx}{\sqrt{1-x^2}}={\left[\frac{x^2}{2}+\frac{x^4}{12}+\frac{3x^6}{60}+\dots \right]}^u$

${\int }_0^u\frac{\sin xdx}{\sqrt{1-x^2}}=\frac{u^2}{2}+\frac{u^4}{12}+\frac{u^6}{20}+\dots$

Thus, the final answer is${\int }_0^u\frac{\sin xdx}{\sqrt{1-x^2}}=\frac{u^2}{2}+\frac{u^4}{12}+\frac{u^6}{20}+\dots$