Question
Question: Integration of tan^-1(cosx/1+sinx)...
Integration of tan^-1(cosx/1+sinx)
(π/4)x - (x^2)/4 + C
Solution
To integrate the given expression, we first simplify the argument of the tan−1 function.
The expression inside tan−1 is 1+sinxcosx.
We can use trigonometric identities to simplify this expression. Recall the identities:
- cosx=sin(2π−x)
- sinx=cos(2π−x)
Substitute these into the expression: 1+sinxcosx=1+cos(2π−x)sin(2π−x) Let A=2π−x. The expression becomes 1+cosAsinA.
Now, use the half-angle formulas for sinA and 1+cosA:
- sinA=2sin(2A)cos(2A)
- 1+cosA=2cos2(2A)
Substitute these into the expression: 1+cosAsinA=2cos2(2A)2sin(2A)cos(2A)=cos(2A)sin(2A)=tan(2A) Now, substitute back A=2π−x: tan(2A)=tan(22π−x)=tan(4π−2x) So, the original integral becomes: ∫tan−1(tan(4π−2x))dx For the principal value of tan−1, if −2π<θ<2π, then tan−1(tanθ)=θ. Assuming x is in a range such that 4π−2x lies within (−2π,2π), we can simplify the integrand: tan−1(tan(4π−2x))=4π−2x Now, we need to integrate this simplified expression: ∫(4π−2x)dx This is a straightforward integration: ∫4πdx−∫2xdx =4πx−21∫xdx =4πx−21(2x2)+C =4πx−4x2+C where C is the constant of integration.