My Questions

1. INT(tan 4^x) dx

2. INT ( (3x+4)/(x^2+1)^2 ) dx

thank you.

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- Mar 24th 2007, 08:08 AM #1

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- Mar 24th 2007, 08:54 AM #2
is it tan(4x) or tan^4(x)

i'll assume the latter since the first would be too easy

int{tan^4(x)}dx

= int{tan^2(x)*tan^2(x)}dx

= int{(sec^2(x) - 1)*tan^2(x)}dx .......trig identity: 1 + tan^2(x) = sec^2(x)

= int{tan^2(x)sec^2(x) - tan^2(x)}dx ...........i multiplied out

= int{tan^2(x)sec^2(x)}dx - int{tan^2(x)}dx ..............split the integral in 2

for the first: int{tan^2(x)sec^2(x)}dx, we proceed by substitution

let u = tan(x)

=> du = sec^2(x) dx

so our integral becomes

int{u^2}du

= (1/3)u^3

= (1/3)tan^3(x)

for the second part: int{tan^2(x)}dx

= int{sec^2(x) - 1}dx .............the same identity

= tan(x) - x

so int{tan^2(x)sec^2(x)}dx - int{tan^2(x)}dx = (1/3)tan^3(x) - tan(x) + x + C

- Mar 24th 2007, 09:04 AM #3

- Mar 24th 2007, 09:05 AM #4

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- Mar 24th 2007, 09:06 AM #5

- Mar 24th 2007, 09:10 AM #6

- Mar 24th 2007, 09:13 AM #7

- Mar 24th 2007, 09:13 AM #8

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- Mar 24th 2007, 09:37 AM #9

- Mar 24th 2007, 09:45 AM #10

- Mar 24th 2007, 12:20 PM #11

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- Mar 24th 2007, 12:22 PM #12

- Mar 24th 2007, 12:26 PM #13

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- Mar 24th 2007, 12:28 PM #14