Integrating Sqrt(1-2sinxcosx): A Short Guide

[gona]

The question is Intigrate $$\sqrt{1-Sin2x}$$

this was my attempt to solve the question

sin2x = 2sinxcosx

(1-2sinxcosx)^1/2

u = (1-2sinxcosx)


$$\frac{du}{dx}$$ = (x-2cos$$^{}2$$x)

du = (x-2cos$$^{}2$$x) dx

im not completely confident about the u substitution rule becouse i have not learned it yet at school. this is as far as i can go with reading in my textbook. is this the correct method to approch this type of question? and if it is what should i do from here?

 

[Gib Z]

$$\int \sqrt{ 1 - \sin (2x)} dx = \int \sqrt{ 1 - 2\sin x \cos x} dx$$
$$= \int \sqrt{ \sin^2 x - 2\sin x \cos x + \cos^2 x} dx$$
$$= \int \sqrt{ ( \sin x - \cos x)^2} dx = \int \sin x - \cos x dx$$
$$= -( \sin x + \cos x) + C$$

 

[christianjb]

Thanks GibZ, that's a good one.

 

[gona]

wow that worksout so nicely thnx alot!

 

[Gib Z]

Quite Proud I made the answer turn out so nicely actually :) The Integrator gives some ugly thing.