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Question Number 40716 by ajeetyadav4370 last updated on 26/Jul/18

$$\int (cosx-cos2x/1-cosx)dx$$

Answered by maxmathsup by imad last updated on 26/Jul/18

$$letI=\int \frac{cosx-cos\left(2x\right)}{1-cosx}dx$$$$I=\int \frac{cosx-(2{cos}^{2}x-1)}{1-cosx}dx=\int \frac{-2{cos}^{2}x+cosx+1}{1-cosx}dx$$$$=\int \frac{1-{cos}^{2}x+cosx-{cos}^{2}x}{1-cosx}dx$$$$=\int \frac{(1-cosx)(1+cosx)+cosx(1-cosx)}{1-cosx}dx$$$$=\int \frac{(1-cosx)(1+2cosx)}{1-cosx}dx=\int (1+2cosx)dx=x+2sinx+k$$$$I=x+2sinx+k.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

$$\int \frac{2sin\frac{3x}{2}sin\frac{x}{2}}{2{sin}^2\frac{x}{2}}dx$$$$\int \frac{sin\frac{3x}{2}}{sin\frac{x}{2}}dx$$$$\int \frac{3sin\frac{x}{2}-4{sin}^3\frac{x}{2}}{sin\frac{x}{2}}dx$$$$\int 3-4{sin}^2\frac{x}{2}dx$$$$3\int dx-2\int (1-cosx)dx$$$$\int dx+2\int cosxdx$$$$x+2sinx+c$$

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