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Question: Integration of tan^-1(cosx/1+sinx)...

Integration of tan^-1(cosx/1+sinx)

Answer

(π/4)x - (x^2)/4 + C

Explanation

Solution

To integrate the given expression, we first simplify the argument of the tan1\tan^{-1} function.

The expression inside tan1\tan^{-1} is cosx1+sinx\frac{\cos x}{1+\sin x}.

We can use trigonometric identities to simplify this expression. Recall the identities:

  1. cosx=sin(π2x)\cos x = \sin(\frac{\pi}{2} - x)
  2. sinx=cos(π2x)\sin x = \cos(\frac{\pi}{2} - x)

Substitute these into the expression: cosx1+sinx=sin(π2x)1+cos(π2x)\frac{\cos x}{1+\sin x} = \frac{\sin(\frac{\pi}{2} - x)}{1+\cos(\frac{\pi}{2} - x)} Let A=π2xA = \frac{\pi}{2} - x. The expression becomes sinA1+cosA\frac{\sin A}{1+\cos A}.

Now, use the half-angle formulas for sinA\sin A and 1+cosA1+\cos A:

  1. sinA=2sin(A2)cos(A2)\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)
  2. 1+cosA=2cos2(A2)1+\cos A = 2\cos^2\left(\frac{A}{2}\right)

Substitute these into the expression: sinA1+cosA=2sin(A2)cos(A2)2cos2(A2)=sin(A2)cos(A2)=tan(A2)\frac{\sin A}{1+\cos A} = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{2\cos^2\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} = \tan\left(\frac{A}{2}\right) Now, substitute back A=π2xA = \frac{\pi}{2} - x: tan(A2)=tan(π2x2)=tan(π4x2)\tan\left(\frac{A}{2}\right) = \tan\left(\frac{\frac{\pi}{2} - x}{2}\right) = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) So, the original integral becomes: tan1(tan(π4x2))dx\int \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) dx For the principal value of tan1\tan^{-1}, if π2<θ<π2-\frac{\pi}{2} < \theta < \frac{\pi}{2}, then tan1(tanθ)=θ\tan^{-1}(\tan \theta) = \theta. Assuming xx is in a range such that π4x2\frac{\pi}{4} - \frac{x}{2} lies within (π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), we can simplify the integrand: tan1(tan(π4x2))=π4x2\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) = \frac{\pi}{4} - \frac{x}{2} Now, we need to integrate this simplified expression: (π4x2)dx\int \left(\frac{\pi}{4} - \frac{x}{2}\right) dx This is a straightforward integration: π4dxx2dx\int \frac{\pi}{4} dx - \int \frac{x}{2} dx =π4x12xdx= \frac{\pi}{4}x - \frac{1}{2} \int x dx =π4x12(x22)+C= \frac{\pi}{4}x - \frac{1}{2} \left(\frac{x^2}{2}\right) + C =π4xx24+C= \frac{\pi}{4}x - \frac{x^2}{4} + C where CC is the constant of integration.