Question

I need help with number 5

Answer 1

Answer:

$\int\limits {{(sin \ x})^{-1} } \, dx = \text{ln}\left |{tan\, \left (\dfrac{x}{2} \right)} \right |$

Step-by-step explanation:

$\int\limits {(sin \ x)^{-1}} \, dx = \int\limits {\dfrac{1}{sin \ x} } \, dx$

We have the following relationships;

$\dfrac{1}{sin \ x } = csc \, x$

We can write;

$csc \, x = csc \, x \times \dfrac{csc \, x + cot \, x}{csc \, x + cot \, x} = \dfrac{csc^2 \, x + csc \, x \cdot cot \, x}{csc \, x + cot \, x}$

We note that the numerator of $\dfrac{csc^2 \, x + csc \, x \cdot cot \, x}{csc \, x + cot \, x}$ , which is ${csc^2 \, x + csc \, x \cdot cot \, x}$ is the derivative of the denominator, ${csc \, x + cot \, x}$, therefore, we can use integration by substitution method and write;

${csc \, x + cot \, x} = u$, from which we get;

$({csc^2 \, x + csc \, x \cdot cot \, x}) \cdot dx = (-1)du$

Therefore, we can write;

$\int\limits {\dfrac{1}{sin \ x} } \, dx = \int\limits {\dfrac{{csc^2 \, x + csc \, x \cdot cot \, x}}{{csc \, x + cot \, x}} } \, dx \Rightarrow -\int\limits {\dfrac{1}{u} } \, du = -ln \left |u \right |$

$\text{-ln} \left |u \right | = \text{-ln}\left |{csc \, x + cot \, x} \right |$

Therefore;

$\int\limits {\dfrac{1}{sin \ x} } \, dx = \text{-ln}\left |{csc \, x + cot \, x} \right |$

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;

$\text{-ln}\left |{csc \, x + cot \, x} \right | = \text{-ln}\left |{cot \, \left (\dfrac{x}{2} \right) } \right | = \text{ln}\left |{cot \, \left (\dfrac{x}{2} \right)} \right | ^{-1} = \text{ln}\left |{tan\, \left (\dfrac{x}{2} \right)} \right |$

Therefore;

$\int\limits {{(sin \ x})^{-1} } \, dx = \int\limits {\dfrac{1}{sin \ x} } \, dx = \text{ln}\left |{tan\, \left (\dfrac{x}{2} \right)} \right |$