   Chapter 8.4, Problem 71E

Chapter
Section
Textbook Problem

# Verifying Formulas Use trigonometric substitution to verify the integration formulas given in Theorem 8.2.

To determine

To prove: The givenintegration formulas with the help of trigonometric substitution.

Explanation

Given:

The 3 integration formulas are:

a2u2du=12(a2arcsinua+ua2u2)+c

u2a2du=12(uu2a2a2ln|u+u2a2|)+c

u2+a2du=12(uu2+a2+a2ln|u+u2+a2|)+c

Proof:

Consider the following first formula:

a2u2du=12(a2arcsinua+ua2u2)+c

Let’s assume u=asinθ;

Now, differentiate above expression with respect to θ, to get;

du=acosθdθ

As u=asinθ;

Then,

a2u2=acosθ

By applying the substitution

a2u2du=acosθacosθdθ

a2u2du=a2cos2θdθ

Now use the trigonometric identity cos2θ=1+cos2θ2 to get;

a2u2du=a2(1+cos2θ2)dθ=a22(1+cos2θ)dθ

Integrate further to get,

a2u2du=a22(θ+12sin2θ)+c;

Now use the trigonometric identity sin2θ=2sinθcosθ to get;

a2u2du=a22(θ+sinθcosθ)+c

We know, u=asinθ

Hence, θ=arcsinua;

So,

a2u2du=a22(arcsinua+(ua)(a2u2a))+c

By simplifying further,

a2u2du=a22(arcsinua+ua2u2a2)+c

Which gives;

a2u2du=12(a2arcsinua+ua2+u2)+c.

(2)

Consider the following second formula,

u2a2du=12(uu2a2a2ln|u+u2a2|)+c

Let’s assume u=asecθ;

By differentiating with respect to θ, we get;

du=asecθtanθdθ

As u=asinθ;

Then,

u2a2=atanθ

By applying the substitution;

u2a2du=atanθasecθtanθdθ=a2tan2θsecθdθ

Now use the trigonometric identity tan2θ=sec2θ1;

u2a2du=a2(sec2θ1)secθdθ=a2(sec3θsecθ)dθ=a2[sec3θdθsecθdθ]

By using the reduction formula secnxdx=secn2xtanxn1+n2n1secn2xdx we get;

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