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### An Integration with Substitution

```
Date: 06/06/98 at 22:05:28
From: Abecina
Subject: Integrations

Hello Dr Math,

I am having a few problems with some integrations. Will you please
help me?

1) Use the result of INT(0 to a) f(x) dx = INT (0 to a) f(a-x) dx to
show that:

x sin(x)        pi^2
INT(0 to pi) ------------ dx = ----
1 + cos^2(x)       4

2) If:

x^n
I(n) = INT(0 to 1) ------- dx
x^2 + 1

1
i) Show that I(n) + I(n-2) = -----
n - 1

ii) Evaluate I(4) and I(5).

Thank you.
```

```
Date: 06/07/98 at 07:53:39
From: Doctor Anthony
Subject: Re: Integrations

Problem 1:

Since sin(pi-x) = sin(x)  and cos(pi-x) = -cos(x), we can use the
above theorem to write the following equation:

x sin(x) dx        (pi-x) sin(x) dx
I = INT ------------ = INT ----------------
1 + cos^2(x)         1 + cos^2(x)

Then adding these two values of I we get:

pi sin(x)dx
2I = INT ------------
1 + cos^2(x)

Let cos(x) = u. Thus -sin(x) dx = du, and x = 0 implies u = 1 and
x = pi implies u = -1. So:

-pi du                  pi du
2I =  INT(1 to -1) ------- = INT(-1 to 1) -------
1 + u^2                1 + u^2

2I = pi tan^(-1)(u) from -1 to 1

= pi [pi/4 - (-pi/4)]

= pi [pi/2]

and so:

I = pi^2/4  as required.

Problem 2:

(i) We have:

x^n + x^(n-2)
I(n) + I(n-2) =  INT [-------------]dx
x^2 + 1

x(n-2)(x^2 + 1)
= INT[----------------]dx
x^2 + 1

= INT[x^(n-2) dx]   from 0 to 1

x^(n-1)
= -------    from 0 to 1
n - 1

1
= -----     as required
n - 1

(ii) Evaluate I(4) and I(5).

I(4) + I(2) = 1/3

I(2) + I(0) = 1/1

and

I(0) = INT[dx/(1+x^2)]  = tan^(-1)(x)  from 0 to 1

=  pi/4

I(2) = 1 - pi/4

I(4) = 1/3 - (1 - pi/4)

I(4) = pi/4 - 1 + 1/3  =  pi/4 - 2/3

For I(5):

I(5) + I(3) = 1/4

I(3) + I(1) = 1/2

and
x dx
I(1) = INT[-------]
1 + x^2

Let 1+x^2 = u. Thus 2x dx = du and x dx = (1/2) du. So:

I(1) = (1/2)INT[du/u]  where x=0 implies u=1 and x=1 implies u=2

= (1/2) ln(u)  = (1/2)[ln(2)- ln(1)]

= (1/2) ln(2)

So I(5) = 1/4 - [1/2 - (1/2)ln(2)]

= (1/2)ln(2) - 1/2 + 1/4

= (1/2)ln(2) - 1/4

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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