question on recurring integrals (1 Viewer)

FD3S-R

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how do i do a question like this

If In = integral (tan^(n)x dx for n greater than/equal to 0, show that In = 1/(n-1)tan^(n-1)x-I(n-2) for n greater than/equal to 2.

Exercise 5.5 q15 i think its cambridge
 

Captain pi

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Break tann(x) into tann – 2(x) and tan²x.

Use the fact that tan²(x) + 1 = sec²(x).

Recognize that tann – 2(x).sec²(x) can be operated on by the reverse chain rule.

Have fun!

Love,

pi.
 

Ogden_Nash

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int{tan<sup>n</sup>x dx} = int{tan<sup>n-2</sup>x * tan<sup>2</sup>x dx}
= int{tan<sup>n-2</sup>x * (sec<sup>2</sup>x - 1) dx}
= int{tan<sup>n-2</sup>x * sec<sup>2</sup>x dx} - int{tan<sup>n-2</sup>x dx}
= tan<sup>n-1</sup>x/n-1 - I<sub>n-2</sub> .... QED

This question is a bit tricky because you don't have to use integration by parts.

[Edit] Nice work Pi, you're too quick.
 

FD3S-R

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bah thanks alot guys, i was trying to use tanx and tan^(n-1)x just got introduced to these questions had no idea....

also any ideas on integral of xsecxtanx...

normally questions are with product of 2 x's but this has 3...
 

Captain pi

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Ogden_Nash said:
int{tan<sup>n</sup>x dx} = int{tan<sup>n-2</sup>x * tan<sup>2</sup>x dx}
= int{tan<sup>n-2</sup>x * (sec<sup>2</sup>x - 1) dx}
= int{tan<sup>n-2</sup>x * sec<sup>2</sup>x dx} - int{tan<sup>n-2</sup>x dx}
= tan<sup>n-1</sup>x/n-1 - I<sub>n-2</sub> .... QED

This question is a bit tricky because you don't have to use integration by parts.

[Edit] Nice work Pi, you're too quick.
You did the hard work; I merely gave a quick overview.
 

Ogden_Nash

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FD3S-R said:
bah thanks alot guys, i was trying to use tanx and tan^(n-1)x just got introduced to these questions had no idea....

also any ideas on integral of xsecxtanx...

normally questions are with product of 2 x's but this has 3...
int{xsecxtanx dx} = xsecx - int{secx dx}
= xsecx - ln|secx + tanx| + C

Remember, derivative of secx = secxtanx and derivative of ln|secx + tanx| = secx
 

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