Jumat, 14 Agustus 2009

Math Integral: Trigonometry

For the upcoming math test, I'll have the instant-kill trigonometry. So here I recap you with trigonometry integrals and other formulas which I think will be useful for solving problems. Original site here.

Basic definition of trigonometric function:
  • $ \displaystyle{ \tan x = { \sin x \over \cos x } } $
  • $ \displaystyle{ \sec x = { 1 \over \cos x } } $
  • $ \displaystyle{ \cot x = { \cos x \over \sin x } = { 1 \over \tan x } } $
  • $ \displaystyle{ \csc x = { 1 \over \sin x } } $
And then:

        • A.) $ \cos^2 x + \sin^2 x = 1 $
        • B.) $ \sin 2x = 2 \sin x \cos x $
        • C.) $ \cos 2x = 2 \cos^2 x - 1 $ so that $ \cos^2 x = \displaystyle{ 1+\cos 2x \over 2}$
        • D.) $ \cos 2x = 1 - 2 \sin^2 x $ so that $ \sin^2 x = \displaystyle{ 1-\cos 2x \over 2} $
        • E.) $ \cos 2x = \cos^2 x - \sin^2 x $
        • F.) $ 1 + \tan^2 x = \sec^2 x $ so that $ \tan^2 x = \sec^2 x - 1 $
        • G.) $ 1 + \cot^2 x = \csc^2 x $ so that $ \cot^2 x = \csc^2 x - 1 $
Basic integrals:

  • $ D (\sin x) = \cos x $
  • $ D (\cos x) = - \sin x $
  • $ D (\tan x) = \sec^2 x $
  • $ D (\cot x) = - \csc^2 x $
  • $ D (\sec x) = \sec x \tan x $
  • $ D (\csc x) = - \csc x \cot x $
Another:

    • 1.) $ \displaystyle{ \int \cos x \, \ dx } \ = \ \sin x + C $
    • 2.) $ \displaystyle{ \int \sin x \, \ dx } \ = \ - \cos x + C $
    • 3.) $ \displaystyle{ \int \sec^2 x \, \ dx } \ = \ \tan x + C $
    • 4.) $ \displaystyle{ \int \csc^2 x \, \ dx } \ = \ - \cot x + C $
    • 5.) $ \displaystyle{ \int \sec x \tan x \, \ dx } \ = \ \sec x + C $
    • 6.) $ \displaystyle{ \int \csc x \cot x \, \ dx } \ = \ - \csc x + C $

That's pretty much for the upcoming exam. Hope can do it at that time -_-".

Pray for me folks!
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