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[Sabine]

please i need this now very urgent

integral (2- (sinx)^2) * (2+ (sinx)^2) / (cos x)^2 please now

 

[monet A]

Hi Sabine what can you do with your double angle formulae here?

$$1- cos2x = 2sin^2x$$
$$1/2(cos2x +1) = cos^2x$$

 

[Sabine]

it can be int(4- sin^2 x)/ cos^2 x dx = int(3+ cos^4 x)\ cos^2 x dx= 3(1+tan^2 x) + 1\tanx

 

[Sabine]

hi monet 2sin^2 x = 1- cos 2x this means what u wrote is wrong .

 

[monet A]

hi monet 2sin^2 x = 1- cos 2x this means what u wrote is wrong .

oops yeah sorry nowhere near a mentor here I just did a quick rearrange of the equation I was looking at to post it probably should have left that part alone, all I really have to offer is that the double angle formulas can really crack open the integral for you, have you applied them?

edit: I am not being real clear am I what I mean is tht I looked at the integral for you because you said its urgent (I'd normally butt out and leave it to the graduates) and I see that if you multiply out the numerator you'll get a simpler integral to manipulate with the double angle formula.

 

[Sabine]

well i tried i jst managed to find the solution i wrote before

 

[geosonel]

integral (2- (sinx)^2) * (2+ (sinx)^2) / (cos x)^2 please now

(2- (sinx)^2) * (2+ (sinx)^2) / (cos x)^2
= (4 - (sinx)^4)/(cos x)^2
= (4 - {(sinx)^2}^2)/(cos x)^2
= (4 - {1 - (cosx)^2}^2)/(cos x)^2
= (4 - {1 - 2(cosx)^2 + (cosx)^4})/(cos x)^2
= (3 + 2(cosx)^2 - (cosx)^4)/(cos x)^2
= (3/(cos x)^2) + 2 - (cos x)^2
= (3/(cos x)^2) - (1/2)*{2(cos x)^2 - 1 - 3}
= (3/(cos x)^2) - (1/2)*{cos(2x) - 3}
= 3(sec x)^2 - (1/2)cos(2x) + (3/2)

integral = 3*tan(x) - (1/4)*sin(2x) + (3/2)*x + C

 

[mathelord]

brilliant geosonel