Find the integrals
Ans fast ??

Solution

I = \int \frac{1}{1 + tanx} dx Now, tanx = \frac{sinx}{cosx} So, I = \int \frac{1}{1 + \frac{sinx}{cosx}} dx = \int \frac{cosx}{cosx + sinx} dx multiplying numerator and denominator by 2, I = \int \frac{2 cosx}{2 (cosx + sinx)} dx = \int \frac{cosx + sinx + cosx - sinx}{2 (cosx + sinx)} dx = \int \frac{1}{2} dx + \frac{1}{2} \int \frac{cosx - sinx}{cosx + sinx} dx = \frac{x}{2} + \frac{1}{2} \int \frac{cosx - sinx}{cosx + sinx} dx taking cosx + sinx = t (cosx - sinx) dx = dt I = \frac{x}{2} + \frac{1}{2} \int \frac{dt}{t} \Rightarrow \frac{x}{2} + \frac{1}{2} \ln |t| + C Now, t = cosx + sinx \Rightarrow I = \frac{x}{2} + \frac{1}{2} \ln |cosx + sinx| + C Hence, I = \frac{x}{2} + \frac{1}{2} \ln | cosx + sinx | + C