Answer to Question #134549 in Calculus for sire

Question

Use Z substitution
I= ∫ dx /(1+sinx)

Answer 1

$\displaystyle\textsf{Let}\hspace{0.1cm}I = \int \frac{1}{1 + \sin{x}}\hspace{0.1cm} \mathrm{d}x \\ \textsf{Substitute}\hspace{0.1cm} Z = \tan\left(\frac{x}{2}\right) \\ \begin{aligned} I &= \int \frac{1}{1 + \frac{2Z}{1 + Z^2}}\hspace{0.1cm} \frac{2\mathrm{d}Z}{1 + Z^2}\\ &= \int \frac{1}{1 + Z^2 + 2Z}\hspace{0.1cm}2\mathrm{d}Z = \int \frac{2}{(Z + 1)^2}\hspace{0.1cm}\mathrm{d}Z \\&= \frac{-2}{1 + Z} + C \\&= \frac{-2}{1 + \tan\left(\frac{x}{2}\right)} + C\\ \end{aligned}\\ \textsf{Multiplying the numerator and}\\ \textsf{denominator} \hspace{0.1cm}\textsf{of}\hspace{0.1cm} I\hspace{0.1cm}\textsf{by} \hspace{0.1cm}\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right), \hspace{0.1cm} \textsf{we have;}\\ I =\frac{-2\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right)} + C\\ \textsf{It is also known that}\hspace{0.1cm} \\\sin{(x)} = 2\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) \hspace{0.1cm}\&\hspace{0.1cm}\\\cos{(x)} = 1 - 2\sin^2\left(\frac{x}{2}\right)\\ \begin{aligned} \therefore I &=\frac{-2\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right)} + C\\ &=\frac{-\sin{x}}{\frac{\sin{x}}{2} + \frac{1 - \cos{x}}{2}} + C= \frac{-2\sin{x}}{1 - \cos{x} + \sin{x}} + C\\&=\frac{2}{\cot{x} -\cosec{x} - 1} + C \end{aligned}\\ \therefore I = \int \frac{1}{1 + \sin{x}}\hspace{0.1cm} \mathrm{d}x = \frac{2}{\cot{x} -\cosec{x} - 1} + C$