Calculate the indefinite integral of f(x)=1/(cos2x + sin^2x)

3 Answers

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neela | High School Teacher | (Level 3) Valedictorian

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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 +


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william1941 | College Teacher | (Level 3) Valedictorian

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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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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