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### Tips on Integrating Trig Functions

```
Date: 02/01/2001 at 00:47:35
From: Karan
Subject: Integration

Dear Math,

I've been trying to integrate this funtion for quite some time now:

dx/(3+4sin x)^2

I've tried to use u = 1/(3+4sinx), but the answer doesn't come out
right.

When I saw the solution, it was solved in a way that required you to
practically know the answer beforehand. They used y = cosx/(3+4sinx),
differentiated it, separated the desired integral and integrated the

Is there any other way? Why isn't my method working?

Thanks,
Karan
```

```
Date: 02/01/2001 at 11:28:12
From: Doctor Rob
Subject: Re: Integration

Thanks for writing to Ask Dr. Math, Karan.

Such tricks are clever, but hard to discover.

There is another way that works whenever you have a rational function
of trigonometric functions, as you do here. The rational function is
1/(3+4*u)^2 and the trigonometric function is u = sin(x).

The systematic method is to use the substitution:

z = tan(x/2).

Then

sin(x) = 2*z/(1+z^2)
cos(x) = (1-z^2)/(1+z^2)
dx = 2*dz/(1+z^2)

Substitute this in, and you'll end up needing to integrate a rational
function of z:

INTEGRAL dx/(3+4*sin[x])^2
= INTEGRAL 1/(3+4*2*z/[1+z^2])^2 * 2*dz/(1+z^2)
= INTEGRAL (1+z^2)^2/(3*z^2+8*z+3)^2 * 2*dz/(1+z^2)
= INTEGRAL 2*(1+z^2)/(3*z^2+8*z+3)^2 * dz

Now you can split this using the fact that

3*z^2 + 8*z + 3 = 3*(z + [4+sqrt(7)]/3)*(z + [4-sqrt(7)]/3)

and partial fractions to break it into parts which you can easily
integrate.

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

```
Date: 02/24/2001 at 14:09:11
From: Karan
Subject: Integration

1. ((tanx)^2+2)^0.5

2. (1-x^2)^n from (0,1) where n is an integer. Do I have to get a
reduction formula? Is there on other way?

Thanks,
Karan
```

```
Date: 02/26/2001 at 09:23:42
From: Doctor Rob
Subject: Re: Integration

Thanks for writing back, Karan.

1. Use the very non-obvious substitution

tan(x) = sqrt(2)*sinh(y)

to reduce this to the integral of 1 + sech(2*y) dy. This may be a bit
easier to deal with.

2. You can avoid the reduction formulas by using the Binomial Theorem
to expand (1-x^2)^n into a sum, and then integrating term by term. I
don't think you can get the integrated sum into a simple closed form
for general n.

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

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