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### Integral Calculus

```
Date: 08/18/99 at 06:20:22
From: Kunal Kulkarni
Subject: Integral calculus

I don't know how to solve integral of sin^10x
```

```
Date: 08/18/99 at 14:57:12
From: Doctor Rob
Subject: Re: Integral calculus

Thanks for writing to Ask Dr. Math.

This can be done using integration by parts.

INTEGRAL sin^n(x)*dx

= INTEGRAL u*dv,

= u*v - INTEGRAL v*du.

Here you should pick u = sin^(n-1)(x)   and   v = -cos(x), so

du = (n-1)*sin^(n-2)(x)*cos(x)*dx   and   dv = sin(x)*dx.

Thus

INTEGRAL sin^n(x)*dx

=   -sin^(n-1)(x)*cos(x)

+ (n-1)*INTEGRAL sin^(n-2)(x)*cos^2(x)*dx,

=   -sin^(n-1)(x)*cos(x)

+ (n-1)*INTEGRAL sin^(n-2)(x)*(1-sin^2(x))*dx,

=   -sin^(n-1)(x)*cos(x)

+ (n-1)*INTEGRAL sin^(n-2)(x)*dx

- (n-1)*INTEGRAL sin^n(x)*dx.

Now bring the least term on the right over to the left and divide
by n:

INTEGRAL sin^n(x)*dx

=  -(1/n)*sin^(n-1)(x)*cos(x) + ((n-1)/n)*INTEGRAL sin^(n-2)(x)*dx

This formula lets you reduce the exponent from 10 to 8, then to 6,
then to 4, then to 2, and finally to INTEGRAL dx, which is easy.

Another approach is to use a trigonometric identity to express
sin^10(x) as a sum of terms of the form

a constant times sine or cosine of a multiple of x

For example,

sin^2(x) = (1/2)(1 - cos(2x))

= (1/2) - (1/2)cos(2x)

sin^3(x) = (1/4)(3sin(x) - sin(3x))

= (3/4)sin(x) - (1/4)sin(3x)

sin^4(x) = (1/8)(3 - 4cos(2x) + cos(4x))

= (3/8) - (1/2)cos(2x) + (1/8)cos(4x)

and so on. These can be derived by using the identities

sin(x)*sin(n*x) = (1/2)[cos([n-1]*x) - cos([n+1]*x)],

sin(x)*cos(n*x) = (1/2)[sin([n-1]*x) + sin([n+1]*x)],

repeatedly for different values of n.

For example, to find sin^5(x), multiply the equation for sin^4(x) on
both sides by sin(x), and use the second of these last two identities,
twice:

sin^5(x) = sin(x)*sin^4(x),

= sin(x)*[(3/8) - (1/2)cos(2x) + (1/8)cos(4x)],

=   (3/8)sin(x)

- (1/2)sin(x)cos(2x)

+ (1/8)sin(x)cos(4x)

=   (3/8)sin(x)

- (1/2)(1/2)[sin([2-1]*x) + sin([2+1]*x)]

+ (1/8)(1/2)[sin([4-1]*x) + sin([4+1]*x)]

=   (3/8)sin(x)

- (1/4)[sin(x) + sin([3x)]

+ (1/16)[sin(3x) + sin(5x)]

=   (3/8)sin(x)

- (1/4)[sin(x) + sin(3x)]

+ (1/16)[sin(3x) + sin(5x)]

=   (3/8)sin(x)

- (1/4)sin(x)

- (1/4)sin(3x)

+ (1/16)sin(3x)

+ (1/16)sin(5x)

=   (1/8)sin(x)

- (3/16)sin(3x)

+ (1/16)sin(5x)

Once you have expressed sin^10(x) that way, then the integration is
easy.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/23/1999 at 04:25:29
From: kunal kulkarni
Subject: Integral calculus

Sir,

Can you also please tell me the integral of 1/x*(1+x^4)^3/4?
```

```
Date: 09/24/1999 at 10:00:31
From: Doctor Rob
Subject: Re: Integral calculus

Assuming that the expression above is the same as

(1/x)*([1+x^4]^3)/4,

and you are integrating with respect to x, the answer may be found by
expanding the cube using the Binomial Theorem,

INTEGRAL (1/x)*(1+3*x^4+3*x^8+x^12)/4 dx,

and expanding the integrand into powers of x,

INTEGRAL (1/4)*x^(-1) + (3/4)*x^3 + (3/4)*x^7 + (1/4)*x^11 dx.

Now the integration is pretty easy, and you can finish.

Another approach would be to let u = 1 + x^4, and so du = 4*x^3 dx,
dx/x = du/(4*[u-1]), and then you want to find

INTEGRAL u^3/(4*[u-1]) du.

This you can do by dividing u - 1 into u^3 by long division, and
finding the quotient and remainder, and then integrating the resulting
expression. This is also pretty easy, and you can do that one
yourself, too.

If instead you meant that the expression to be integrated to be

(1/x)*(1+x^4)^(3/4),

then this is one of those functions that cannot be integrated in
closed form in terms of the familiar functions of calculus. My version
of Mathematica(TM) gives the result:

(1+x^4)^(3/4)/3 - Hypergeometric2F1[1/4,1/4,5/4,-x^(-4)]/Abs[x].

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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