C4 IntegrationWatch

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#1
integrate 1/(sin^2 x.cos^2 x)

0
13 years ago
#2
Using the identity sin(2x) = 2sin(x)cos(x),

1 / (sin^2(x)cos^2(x))
= 1 / ((1/2)sin(2x))^2
= 4cosec^2(2x)

(int) 1 / (sin^2(x)cos^2(x)) dx
= (int) 4cosec^2(2x) dx
= (int) 2cosec^2(u) du . . . . . u = 2x, du/dx = 2
= -2cot(u) + c . . . . . using the formula booklet
= -2cot(2x) + c
0
#3
(Original post by Jonny W)
1 / (sin^2(x)cos^2(x))
= 1 / ((1/2)sin(2x))^2
= 4cosec^2(2x)

(int) 1 / (sin^2(x)cos^2(x)) dx
= (int) 4cosec^2(2x) dx
= (int) 2cosec^2(u) du . . . . . u = 2x, du/dx = 2
= -2cot(u) + c . . . . . using the formula booklet
= -2cot(2x) + c
in the second method i see you used the substitution method, what did you do in the first example exactly?

thanks!
0
13 years ago
#4
I = ∫ 1/(sinxcosx)Â² dx
= ∫ 1/(0.5sin(2x))Â² dx
= ∫ (2cosec(2x))Â² dx
= ∫ 4cosecÂ²(2x) dx
= -4 ∫ -cosecÂ²(2x) dx
= -4 (1/2) cot(2x) + C
= -2 cot(2x) + C

Or:
I = ∫ (sinÂ²x + cosÂ²x)/(sinxcosx)Â²
= ∫ (sinÂ²x)/(sinxcosx)Â² dx + ∫ (cosÂ²x)/(sinxcosx)Â²
= ∫ 1/cosÂ²x dx + ∫ 1/sinÂ²x dx
= ∫ secÂ²x dx + ∫ cosecÂ²x dx
= tanx - cotx + C
0
#5
(Original post by dvs)
I = ∫ 1/(sinxcosx)Â² dx
= ∫ 1/(0.5sin(2x))Â² dx
= ∫ (2cosec(2x))Â² dx
= ∫ 4cosecÂ²(2x) dx
= -4 ∫ -cosecÂ²(2x) dx
= -4 (1/2) cot(2x) + C
= -2 cot(2x) + C

Or:
I = ∫ (sinÂ²x + cosÂ²x)/(sinxcosx)Â²
= ∫ (sinÂ²x)/(sinxcosx)Â² dx + ∫ (cosÂ²x)/(sinxcosx)Â²
= ∫ 1/cosÂ²x dx + ∫ 1/sinÂ²x dx
= ∫ secÂ²x dx + ∫ cosecÂ²x dx
= tanx - cotx + C
ah sorry guys- i see what you've done- used trig (additon formulae)!!!!!
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