$\int \frac{1 + sin x}{sin x \cdot (1 + cos x)} d x$ $int (1+sinx)/(sinx * (1+cosx))dx$ ?

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$\int \frac{1 + sin x}{sin x \cdot (1 + cos x)} d x$ $int (1+sinx)/(sinx * (1+cosx))dx$ ?

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Answer 1 · 2018-04-03T18:09:15

$\frac{1}{4} {(tan (\frac{x}{2}))}^{2} + tan (\frac{x}{2}) + \frac{1}{2} ln (tan (\frac{x}{2})) + C$ $1/4(tan(x/2))^2+tan(x/2)+1/2Ln(tan(x/2))+C$

Explanation:

$\int \frac{1 + sin x}{sin x \cdot (1 + cos x)} \cdot d x$ $int (1+sinx)/(sinx*(1+cosx))*dx$

After using $y = tan (\frac{x}{2})$ $y=tan(x/2)$ , $d x = \frac{2 d y}{y^{2} + 1}$ $dx=(2dy)/(y^2+1)$ , $sin x = \frac{2 y}{y^{2} + 1}$ $sinx=(2y)/(y^2+1)$ and $cos x = \frac{1 - y^{2}}{y^{2} + 1}$ $cosx=(1-y^2)/(y^2+1)$ transforms, this integral became

$\int \frac{1 + \frac{2 y}{y^{2} + 1}}{\frac{2 y}{y^{2} + 1} (1 + \frac{1 - y^{2}}{y^{2} + 1})} \cdot \frac{2 d y}{y^{2} + 1}$ $int (1+(2y)/(y^2+1))/((2y)/(y^2+1)(1+(1-y^2)/(y^2+1)))*(2dy)/(y^2+1)$

= $\int \frac{\frac{y^{2} + 2 y + 1}{y^{2} + 1}}{\frac{2 y}{y^{2} + 1} \frac{2}{y^{2} + 1}} \cdot \frac{2 d y}{y^{2} + 1}$ $int ((y^2+2y+1)/(y^2+1))/((2y)/(y^2+1)2/(y^2+1))*(2dy)/(y^2+1)$

= $\int \frac{(y^{2} + 2 y + 1) \cdot d y}{2 y}$ $int ((y^2+2y+1)*dy)/(2y)$

= $\frac{1}{2} \int y \cdot d y + \int d y + \frac{1}{2} \int \frac{d y}{y}$ $1/2int y*dy+int dy+1/2int (dy)/y$

= $\frac{y^{2}}{4} + y + \frac{1}{2} ln y + C$ $y^2/4+y+1/2Lny+C$

= $\frac{1}{4} {(tan (\frac{x}{2}))}^{2} + tan (\frac{x}{2}) + \frac{1}{2} ln (tan (\frac{x}{2})) + C$ $1/4(tan(x/2))^2+tan(x/2)+1/2Ln(tan(x/2))+C$

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$\int \frac{1 + sin x}{sin x \cdot (1 + cos x)} d x$ $int (1+sinx)/(sinx * (1+cosx))dx$ ?

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Explanation:

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∫1+sinxsinx⋅(1+cosx)dxint (1+sinx)/(sinx * (1+cosx))dx ?

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$\int \frac{1 + sin x}{sin x \cdot (1 + cos x)} d x$ $int (1+sinx)/(sinx * (1+cosx))dx$ ?