Answer:
[tex]\int\limits {{(sin \ x})^{-1} } \, dx = \text{ln}\left |{tan\, \left (\dfrac{x}{2} \right)} \right |[/tex]
Step-by-step explanation:
[tex]\int\limits {(sin \ x)^{-1}} \, dx = \int\limits {\dfrac{1}{sin \ x} } \, dx[/tex]
We have the following relationships;
[tex]\dfrac{1}{sin \ x } = csc \, x[/tex]
We can write;
[tex]csc \, x = csc \, x \times \dfrac{csc \, x + cot \, x}{csc \, x + cot \, x} = \dfrac{csc^2 \, x + csc \, x \cdot cot \, x}{csc \, x + cot \, x}[/tex]
We note that the numerator of [tex]\dfrac{csc^2 \, x + csc \, x \cdot cot \, x}{csc \, x + cot \, x}[/tex] , which is [tex]{csc^2 \, x + csc \, x \cdot cot \, x}[/tex] is the derivative of the denominator, [tex]{csc \, x + cot \, x}[/tex], therefore, we can use integration by substitution method and write;
[tex]{csc \, x + cot \, x} = u[/tex], from which we get;
[tex]({csc^2 \, x + csc \, x \cdot cot \, x}) \cdot dx = (-1)du[/tex]
Therefore, we can write;
[tex]\int\limits {\dfrac{1}{sin \ x} } \, dx = \int\limits {\dfrac{{csc^2 \, x + csc \, x \cdot cot \, x}}{{csc \, x + cot \, x}} } \, dx \Rightarrow -\int\limits {\dfrac{1}{u} } \, du = -ln \left |u \right |[/tex]
[tex]\text{-ln} \left |u \right | = \text{-ln}\left |{csc \, x + cot \, x} \right |[/tex]
Therefore;
[tex]\int\limits {\dfrac{1}{sin \ x} } \, dx = \text{-ln}\left |{csc \, x + cot \, x} \right |[/tex]
csc x + cot x = (1/sin x) + ((cos x)/(sin x)) = (1 + cos x)/(sin x)
(1 + cos x)/(sin x) = (cos²(x/2) + sin²(x/2) + cos²(x/2) - sin²(x/2))/(2sin(x/2)·cos(x/2)) = (2·cos²(x/2))/((2sin(x/2)·cos(x/2)) = cos(x/2)/sin(x/2) = cot(x/2)
Therefore;
[tex]\text{-ln}\left |{csc \, x + cot \, x} \right | = \text{-ln}\left |{cot \, \left (\dfrac{x}{2} \right) } \right | = \text{ln}\left |{cot \, \left (\dfrac{x}{2} \right)} \right | ^{-1} = \text{ln}\left |{tan\, \left (\dfrac{x}{2} \right)} \right |[/tex]
Therefore;
[tex]\int\limits {{(sin \ x})^{-1} } \, dx = \int\limits {\dfrac{1}{sin \ x} } \, dx = \text{ln}\left |{tan\, \left (\dfrac{x}{2} \right)} \right |[/tex]
