Very Difficult Trig Sub?

Homework Statement

Integral of 1/(2+sin(x)) dx

The Attempt at a Solution

I've been told that you can use trig subs, but I never had to learn that in high school and it hasn't appeared in any of my calculus coursework.

As a side note. I've been wondering if it is possible to solve asin(x) + bcos(x) = c

To solve your side note use this website. it helped me out! good luck

http://www.education2000.com/demo/demo/btnchtml/sinplcos.htm [Broken]

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Wow, that's so cool. Thanks for the link. I'll try to remember how to derive it.

dextercioby
Homework Helper

Homework Statement

Integral of 1/(2+sin(x)) dx

The standard substitution $\tan\frac{x}{2} = t$ applies for your integral. It will convert it into a integral of an algebraic function for which the method of partial fraction decomposition will get it solved.

Thanks, I'll try that. Is that something you just memorized or is there a certain rule that lets you know what to substitute?

dextercioby
Homework Helper
There are rules. That substitution will apply to an antiderivative of a function a+b\sin x/c+d\cos x and the other 3 ways of interchanging cos with sin and more generally to any algebraic function of sin and cos.

Where can I learn all these rules? I usually only see substitutions with x = asint, atant, or asect

dextercioby
Homework Helper
You're normally taught these rules of substitution in high-school. I wasn't, so I picked them up for myself from books, especially for engineers, because the proofs are missing :)

Okay so, given:
integral dx/(2+sinx)

tan(x/2) = t
(1/2)sec^2 (x/2) dx = dt
dx = 2cos^2 (x/2) dt

integral
2cos^2 (x/2) dt / (2+sinx)

Am I supposed to use x = arctan(2t)? If so, is it possible to simplify by drawing a triangle?

dextercioby
Homework Helper
Of course you have to use that. It's the whole purpose of substitution, you need to change every function of x including the dx with the approproate function of t and dt.

I know how to change sinx to sin 2t/sqrt(1+4t^2)
but I'm not sure how to simplify cos^2 (x/2) since it has the 1/2 in front of the x and I can't use the same trick that I used for sinx.

dextercioby
Homework Helper
But you need sin (2 arctan t) from the initial integral.

$$\sin (2\arctan t) = 2 (\sin\arctan t) (\cos\arctan t)$$

$$\sin\arctan t = \frac{t}{\sqrt{1+t^2}} \, , \, \cos\arctan t = \frac{1}{\sqrt{1+t^2}}$$

What about the integration element ?

Char. Limit
Gold Member
Okay so, given:
integral dx/(2+sinx)

tan(x/2) = t
(1/2)sec^2 (x/2) dx = dt
dx = 2cos^2 (x/2) dt

integral
2cos^2 (x/2) dt / (2+sinx)

Am I supposed to use x = arctan(2t)? If so, is it possible to simplify by drawing a triangle?

It's not x=arctan(2t), but rather x=2arctan(t). that might help.

Thank you. I modified the integral to
dt/t^2+t+1)
Are you sure it's partial fractions?

Char. Limit
Gold Member
There, I would actually use completing the square in the denominator, then do another trig sub.

Thank you. I finally get it now. I'll still have problems with the initial trig substitutions though since I'm not sure how to get tan(x/2) = t.

Letting $t = \tan(x/2)$ is part of something called a Weierstrass substitution. This is usually a pretty messy substitution, but it's good to have in your toolbox of integration tricks, especially for those pesky integrals where nothing else seems to work.