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How do you integrate the following:

i) 1/(4 - x^2) .dx

ii) tanx .dx

i) 1/(4 - x^2) .dx

ii) tanx .dx

Bezza

1) 1/(4-x^2) is difference of 2 squares so equals 1/[(2+x)(2-x)], you then use partial fractions and integrate.

2) tanx = sinx/cosx The derivative of cosx is -sinx so use the substitution u = cosx

2) tanx = sinx/cosx The derivative of cosx is -sinx so use the substitution u = cosx

1) 1 = A(2 - x) + B(2 + x)

x = 2 1=4B so B = 1/4

x = -2 1=4A so A = 1/4

1/4(2 + x) + 1/4(2 - x)

INT = 1/4(2x + x^2/2) + 1/4(2x - x^2/2) + c

2) let u = cosx

du/dx = -sinx

dx = du/-sinx

{sinx/u.1/-sinx .du = -1/u.du = -LNU + c

replace u with cosx

= -LNCOSX + c

are these right?

INT tanx dx = ln |sec x|... its in the formula book, page 8

The first one you've integrated the wrong thing. You need to integrate 1/[4(2-x)] + 1/[4(2+x)]

The second one is right, but is usually simplified to ln|secx| because of the negative sign

As pixelfairy says, both of these general forms are in the formula book so you can check you have the right answer or try and get some clues where to go.

The second one is right, but is usually simplified to ln|secx| because of the negative sign

As pixelfairy says, both of these general forms are in the formula book so you can check you have the right answer or try and get some clues where to go.

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