# What is the antiderivative of f(x)*cos x if f(x)=ln(1+sin^2x)?

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We have to find the anti derivative of f(x)*cos x , where f(x) = ln(1 + (sin x)^2)

Int [ f(x)*cos x dx]

=> Int [ ln(1 + (sin x)^2) * cos x dx]

let sin x = y

dy = cos x dx

=> Int [ ln(1 + y^2) dy]

Use integral by parts.

Let u = ln( y^2 + 1) , du = [ 1 / y^2 + 1)]*2y dy

dv = dy , v = y

so

integral [ ln (y^2 + 1) dy]

=> y ln( y^2 + 1) - Int [2y^2 / (y^2 + 1) dy]

=> y*ln(y^2 + 1) - 2 Int[ 1 - ( 1 / y^2 + 1) dy]

=> y*ln(y^2 + 1) - 2( y - arc tan y) + c

substitute y = sin x

=> sin x*ln((sin x)^2 + 1) - 2*(sin x - arc tan (sin x)) + C

**The antiderivative of the given expression is sin x*ln((sin x)^2 + 1) - 2*(sin x - arc tan (sin x)) + C**

To determine the antiderivative of the product f(x)*cos x, we'll have to determine the indefinite integral of f(x)*cos x.

Int f(x)*cos x dx = Int ln[1+(sin x)^2]*cos x dx

We'll replace sin x by t:

sin x = t

We'll differentiate both sides:

cos x dx = dt

We'll re-write the integral:

Int ln[1+(t)^2]dt

We'll integrate by parts using the formula:

Int udv = u*v - Int vdu

u = ln(1+t^2) => du = 2t dt/(1+t^2)

dv = dt => v = t

Int ln[1+(t)^2]dt = t*ln(1+t^2) - Int 2t^2 dt/(1+t^2)

Int 2t^2 dt/(1+t^2) = 2 Int dt - 2 Int dt/(1+t^2)

Int 2t^2 dt/(1+t^2) = 2t - 2 arctan t + C

The antiderivative of the given function is:

Int ln[1+(t)^2]dt = t*ln(1+t^2) - 2t + 2 arctan t + C

**Int ln[1+(sin x)^2]*cos x dx =sin x*ln[1+(sin x)^2] - 2sin x + 2*arctan (sin x) + C**