P3 Integration Watch

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Aristotle
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#1
Report Thread starter 13 years ago
#1
1) Use the identity sin(A+B) + sin(A-B) = 2sinAcosB to find the integral of:

a) 2sin6xcos4x dx.

b) sinxcos0.5x dx.

2) Find the integral of:

a) sec^4 x dx.

b) cot^4 x dx.

Answers:

1a) -0.1cos10x-0.5cos2x.
1b) -1/3 cos1.5x-cos0.5x.

2a) 1/3 tan^3 x + tanx.
2b) x + cotx - 1/3 cot^3 x.

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m:)ckel
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#2
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1) These are just a matter of actually using the identity given. Very little thinking needed. Just look at the numbers in place of 'A' and 'B' in the given identity, and then it's literally a matter of copying the corresponding numbers of A and B.

a) A=6, B=4..

=> INT sin(6+4)x + sin(6-4)x dx

= INT sin10x + sin2x dx

= (-1/10)cos10x - (1/2)cos2x + c

Exactly the same method for b)
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Fermat
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#3
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#3
2a)
I = int sec^4 x dx.

sec²x = 1 + tan²
sec^4x = sec²x.sec²x = sec²(1+tan²x)
sec^4x = sec²x + sec²x.tan²x
I = int sec²x + sec²x.tan²x dx

int sec²x dx = tanx

you should recognise that sec²x is the differential coefficient of tanx, from which you can see that

int sec²x.tan²x dx = (1/3)tan^3x

So I = tanx + (1/3)tan^3x + c
=======================

you should be able to do part b) now using a similar method and trig identities involving cot²x
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idiopathic
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#4
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Fermat (or anyone) could you show the workings for 2 b) please? I used your advice, but I'm stuck.
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Fermat
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#5
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(Original post by endeavour)
Fermat (or anyone) could you show the workings for 2 b) please? I used your advice, but I'm stuck.
I = int cot^4x dx

cot²x = cosec²x - 1
cot^4x = cot²x.cot²x = cot²x(cosec²x - 1)
cot^4x = cot²x.cosec²x - cot²x
cot^4x = cot²x.cosec²x - cosec²x + 1

I = int cot^4x dx
I = int cot²x.cosec²x - cosec²x + 1 dx

you should recognise that -cosec²x is the differential coefficient of cotx, from which you can see that

int cot²x.cosec²x dx = -(1/3)cot^3x

So, I = -(1/3)cot^3x + cotx + x + c
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