◆50 points questuon◆ solve only on copy.....
Answers
Answer:
Step-by-step explanation:
1 ÷ (cos^2-1)×cos^2= 1/cos^4-cos^2.
To integrate 1/(cos2x+sin^2x).
We shall have a transformation; tanx = t. Thhen sec^x*dx = dt or dx = dt/(1+tan^2x) = dt/(1+t^2.
We know that cos2x = 2cos^2x-1 = 2/(1+t^2) -1 = (1-t^2)/(1+t^2)
sin^2x = t^2/(1+t^2).
Therefore Integral dx/(cos2x+sin^2x) = integral dt/{(1+t^2) (t^2/(1+t^2 +(1-t^2)/(1+t^2)} = Integral dt/{t^2+1-t^2} = Integral dt/1 = t +C
Therefore dx/(cos2x+sin^2x) = t +C = tanx +
william1941 | Student
We have to find the integral of f(x) =1/(cos2x + sin^2x).
We first simplify the denominator using the property for cos 2x which is cos 2x = (cos x)^2 - (sin x)^2.
Therefore f(x) =1 / (cos 2x + sin^2x)
=> f(x) = 1/ [ (cos x)^2 - (sin x)^2 + (sin x)^2 ]
Subtracting (sin x)^2 we get
=> f(x) = 1/ (cos x)^2
=> f(x) = (sec x)^2, as 1/ cos x = sec x.
The integral of (sec x)^2 is tan x.
Therefore the integral of f(x)=1/(cos2x + sin^2x) is tan x+C.
giorgiana1976 | Student
To determine the result of the indefinite integral, we'll have to re-write the denominator. We'll apply the formula of the cosine of a double angle.
cos 2x = cos(x+x) = cosx*cosx - sinx*sinx
cos 2x = (cosx)^2 - (sinx)^2
If we'll pay attention to the terms of the denominator, we'll notive that beside cos 2x, we'll have also the term (sinx)^2. So, we'll re-write cos 2x, with respect to the function sine only.
We'll substitute (cosx)^2 by the difference 1-(sinx)^2:
cos 2x = 1-(sinx)^2 - (sinx)^2
cos 2x = 1-2(sinx)^2
The denominator will become:
cos2x + (sinx)^2 = 1-2(sinx)^2 + (sinx)^2
cos2x + (sinx)^2 = 1-(sinx)^2
But, 1-(sinx)^2 = (cosx)^2 (from the fundamental formula of trigonometry)
cos2x + (sinx)^2 = (cosx)^2
The indefinite integral of f(x) will become:
Int f(x)dx = Int dx/(cosx)^2 = tan x + C