   Chapter 8.6, Problem 49E

Chapter
Section
Textbook Problem

# Verifying a Formula In Exercises 49-54, verify the integration formula. ∫ 1 ( u 2 ± a 2 ) 3 / 2 d u = ± u a 2 u 2 ± a 2 + C

To determine

To prove: The integral formula 1(u2±a2)32du=±ua2u2±a2+C.

Explanation

Given:

The provided integral formula is 1(u2±a2)32du=±ua2u2±a2+C.

Formula used:

The formula of trigonometry is used,

secA=1cosA,tanA=sinAcosA

And Pythagoras theorem

h2=p2+b2

Where h is the hypotenuse, p is the height, and the b is the base.

Proof:

In the integral’s left-handside,

1(u2±a2)32du

For u2+a2, it is simplified by substituting into trigonometric terms,

u=atanθdu=asec2θdθu2+a2=a2sec2θ

Applying integral,

1(u2+a2)32du=asec2θdθa3sec3θ=1a2cosθdθ=1a2sinθ+C

From using Pythagoras theorem,

It is found that,

sinθ=ua2u2+a2+C</

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