Evaluate $\displaystyle \int{\dfrac{\cos{x}-\cos{2x}}{1-\cos{x}}\,}dx$

In this indefinite integral problem, $x$ is a variable but it represents the angle of a right triangle. It formed the following trigonometric function in cosine terms in rational function.

$\dfrac{\cos{x}-\cos{2x}}{1-\cos{x}}$

We have to evaluate the indefinite integral of this trigonometric function in rational form with respect to $x$ in this calculus problem.

$\displaystyle \int{\dfrac{\cos{x}-\cos{2x}}{1-\cos{x}}\,}dx$

Simplify the Trigonometric function

In the trigonometric rational expression, there are two cosine function and one cosine function with double angle. The cosine functions cannot be simplified but the cosine function that contains double angle can be expanded by the cos double angle identity. It makes the entire mathematical expression to express in cosine functions purely.

$=\,\,\,$ $\displaystyle \int{\dfrac{\cos{x}-\Big(2\cos^2{x}-1\Big)}{1-\cos{x}}\,}dx$

Now, simplify the trigonometric expression in the numerator of the rational function.

$=\,\,\,$ $\displaystyle \int{\dfrac{\cos{x}-2\cos^2{x}+1}{1-\cos{x}}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{-\Big(2\cos^2{x}-\cos{x}-1\Big)}{1-\cos{x}}\,}dx$

Factorize the Trigonometric expression

The trigonometric expression in the numerator is in quadratic expression form. So, it can be factored for simplifying the trigonometric expression further.

$=\,\,\,$ $\displaystyle \int{\dfrac{-\Big(2\cos^2{x}-2\cos{x}+\cos{x}-1\Big)}{1-\cos{x}}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{-\Big(2\cos{x}(\cos{x}-1)+1 \times \cos{x}-1 \times 1\Big)}{1-\cos{x}}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{-\Big(2\cos{x}(\cos{x}-1)+1(\cos{x}-1)\Big)}{1-\cos{x}}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{-(\cos{x}-1)(2\cos{x}+1)}{1-\cos{x}}\,}dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{(1-\cos{x})(2\cos{x}+1)}{1-\cos{x}}\,}dx$

$=\,\,\,$ $\require{cancel} \displaystyle \int{\dfrac{\cancel{(1-\cos{x})}(2\cos{x}+1)}{\cancel{1-\cos{x}}}\,}dx$

$=\,\,\,$ $\displaystyle \int{(2\cos{x}+1)}\,dx$

Evaluate the Indefinite Integral of function

The trigonometric expression in rational form is simplified successfully. The integral of sum of terms can be evaluated by the sum rule of integration.

$=\,\,\,$ $\displaystyle \int{2\cos{x}}\,dx+\int{1}\,dx$

Now, use the constant multiple rule of integration for separating the constant $2$ from the first integral term.

$=\,\,\,$ $\displaystyle 2\int{\cos{x}}\,dx+\int{1}\,dx$

Finally, use the integral rule of sine function and integral rule of constant for evaluating the integration of each term in the expression.

$=\,\,\,$ $\displaystyle 2(\sin{x}+c_1)+x+c_2$

Now, focus on simplifying the mathematical expression.

$=\,\,\,$ $\displaystyle 2\sin{x}+2c_1+x+c_2$

$=\,\,\,$ $\displaystyle 2\sin{x}+x+2c_1+c_2$

$=\,\,\,$ $\displaystyle 2\sin{x}+x+c$

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