Chapter 1: Q36q (page 32)

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

Short Answer

Expert verified

$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1 - x^{2}}} = \frac{u^{2}}{2} + \frac{u^{4}}{12} + \frac{u^{6}}{20} + \dots$

Step by step solution

Given information

The Maclaurin expansion of the function $\int_{0}^{u} \frac{\sin x d x}{\sqrt{1 - x^{2}}}$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

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

$s i n (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + \dots$

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

State the Maclaurin series.

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

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

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

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

Use the defined to series and evaluate.

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

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

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

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

Integrate the series.

$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1 - x^{2}}} = \int_{0}^{u} [x + \frac{x^{3}}{3} + \frac{3 x^{5}}{10} + \dots] d x$

$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1 - x^{2}}} = {[\frac{x^{2}}{2} + \frac{x^{4}}{12} + \frac{3 x^{6}}{60} + \dots]}^{u}$

$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1 - x^{2}}} = \frac{u^{2}}{2} + \frac{u^{4}}{12} + \frac{u^{6}}{20} + \dots$

Thus, the final answer is $\int_{0}^{u} \frac{\sin x d x}{\sqrt{1 - x^{2}}} = \frac{u^{2}}{2} + \frac{u^{4}}{12} + \frac{u^{6}}{20} + \dots$

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

Short Answer

Step by step solution

Given information

Definition of Maclaurin series

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

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

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

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∫0usinxdx1-x2

Short Answer

Step by step solution

Given information

Definition of Maclaurin series

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

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

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Scientific Method Physics

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Study anywhere. Anytime. Across all devices.

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