/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q12.2P Use a graph of sin2θ and the t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a graph of sin2θand the text discussion just before (12.4)to verify the equations (12.4). Note that the area under thesin2θ graph from 0-π2and the area from π2-π are mirror images of each other, and this will be true also for any function ofsin2θ.

Short Answer

Expert verified

The statement is verified.

Step by step solution

01

Given Information

The function is sin2θ.

02

Definition of elliptic form.

The elliptic form of the integral is defined asE(π0,k)=∫0π21-k2sin2θdθ.

03

Verify the statement.

The function is sin2θ.

lnπ±ϕ=∫0nπ±ϕfsin2θdθ0≤∅≤π2=∫0nπfsin2θdθ+∫nπnπ±ϕfsin2θdθ=n∫0πf(sin2θ)dθ+∫0ϕf(sin2θ)dθ

Find the value ofn∫0πf(sin2θ)dθ

∫0πf(sin2θ)dθ=∫0π2fsin2θdθ+∫π2πfsin2θdθ=∫π20fsin2π-θ(-dθ)=∫0π2fsin2θdθ=2∫0π2fsin2θdθ

Hence, lnπ±ϕ=2nInπ2±lϕ.

The statement is verified.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.