Question

How do I evaluate $\int {\dfrac{{1 + \cos x}}{{\sin x}}dx}$?

Answer 1

First, separate the denominator in the integral and use trigonometric identities to simplify it. Then, use the property that the integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions. Next, use integral formula for cosecant and cotangent function. Then, use logarithm property and trigonometry identity to further simplify the result.and get the desired result.

Formula used:
The integral of the product of a constant and a function = the constant $\times$ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx}$, where $k$ is a constant.
Trigonometric identity: $\cos ecx = \dfrac{1}{{\sin x}}$, $\cot x = \dfrac{{\cos x}}{{\sin x}}$
The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx}$
Integration formula: $\int {\cos ecxdx} = \log \left| {\cos ecx - \cot x} \right| + C$ and $\int {\cot xdx} = \log \left| {\sin x} \right| + C$

Complete step by step answer:
We have to find $\int {\dfrac{{1 + \cos x}}{{\sin x}}dx}$…(i)
First, separate the denominator in integral (i).
$\int {\left[ {\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}} \right]dx}$…(ii)
Now, use trigonometry identities $\cos ecx = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ in integral (ii).
$\int {\left[ {\cos ecx + \cot x} \right]dx}$…(iii)
Now, using the property that the integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx}$
So, in above integral (iii), we can use above property
$\int {\cos ecxdx} + \int {\cot xdx}$…(iv)
Now, use property $\int {\cos ecxdx} = \log \left| {\cos ecx - \cot x} \right| + C$ and $\int {\cot xdx} = \log \left| {\sin x} \right| + C$ in integral (iv).
$\Rightarrow \log \left| {\cos ecx - \cot x} \right| + \log \left| {\sin x} \right| + C$
Now, use property $\log m + \log n = \log mn$ to simplify the above result.
$\Rightarrow \log \left| {\left( {\cos ecx - \cot x} \right)\left( {\sin x} \right)} \right| + C$
Multiply $\sin x$ to $\left( {\cos ecx - \cot x} \right)$, we get
$\Rightarrow \log \left| {\cos ecx \times \sin x - \cot x \times \sin x} \right| + C$
Now, use trigonometry identities $\cos ecx = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
$\Rightarrow \log \left| {\dfrac{1}{{\sin x}} \times \sin x - \dfrac{{\cos x}}{{\sin x}} \times \sin x} \right| + C$
Cancel the common factor.
$\Rightarrow \log \left| {1 - \cos x} \right| + C$

Hence, $\int {\dfrac{{1 + \cos x}}{{\sin x}}dx} = \log \left| {1 - \cos x} \right| + C$.

Note: $\int {\cot xdx} = \log \left| {\sin x} \right| + C$
We have $\int {\cot xdx} = \int {\dfrac{{\cos x}}{{\sin x}}dx}$
Put $\sin x = t$ so that $\cos xdx = dt$.
Then, $\int {\cot xdx} = \int {\dfrac{{dt}}{t}} = \log \left| t \right| + C = \log \left| {\sin x} \right| + C$.
And, $\int {\cos ecxdx} = \log \left| {\cos ecx - \cot x} \right| + C$
We have $\int {\cos ecxdx} = \int {\dfrac{{\cos ecx\left( {\cos ecx + \cot x} \right)}}{{\left( {\cos ecx + \cot x} \right)}}dx}$
Put $\cos ecx + \cot x = t$ so that $- \cos ecx\left( {\cos ecx + \cot x} \right)dx = dt$.
So, $\int {\cos ecxdx} = - \int {\dfrac{{dt}}{t}} = \log \left| t \right| = - \log \left| {\cos ecx + \cot x} \right| + C$
It can also be written as
$\int {\cos ecxdx} = - \log \left| {\dfrac{{\cos e{c^2}x - {{\cot }^2}x}}{{\cos ecx - \cot x}}} \right| + C$
Therefore, $\int {\cos ecxdx} = \log \left| {\cos ecx - \cot x} \right| + C$.