Question

How do you find the antiderivative of $y=\dfrac{\sin x}{1-\cos x}$ ?

Answer 1

Hint: We have been given a trigonometric function in the fractional format consisting of sine and cosine functions whose antiderivative is to be found. From our prior knowledge of basic calculus, we know that the antiderivative is another name for the integral of the function. Thus, we shall integrate the given trigonometric expression by using simple substitution to simplify the given expression for integration.

Complete step by step solution:
Given that $y=\dfrac{\sin x}{1-\cos x}$.
In order to find the antiderivative of the function given and calculate its derivative, we shall substitute the denominator of the expression equal to some variable-t.
Let $t=1-\cos x$
Differentiating left hand side with respect to variable-t and right hand side with respect to variable-x, we get
$dt=\dfrac{d}{dx}\left( 1-\cos x \right)$
We know that $\dfrac{d}{dx}\cos x=-\sin x$ and the differential of a constant term is zero.
$\Rightarrow dt=0-\left( -\sin x.dx \right)$
$\Rightarrow dt=\sin x.dx$
We shall now find the integral, $I$ of the given expression.
$\Rightarrow I=\int{\dfrac{\sin x}{1-\cos x}}.dx$
Now, substituting the values of variable-t and dt, we have
$\Rightarrow I=\int{\dfrac{dt}{t}}$
We know that $\int{\dfrac{1}{u}.du=\ln u+C}$.
 $\Rightarrow I=\ln t+C$
Further, substituting the value of variable-t, we get
$\Rightarrow I=\ln \left( 1-\cos x \right)+C$
Therefore, the antiderivative of $y=\dfrac{\sin x}{1-\cos x}$ is $\ln \left( 1-\cos x \right)+C$.

Note: Since we do not have a direct formula of integration for the given expression of sine and cosine function, therefore, we integrated this function by applying simple substitution of $t=1-\cos x$ Here, one mistake which we could have made was writing the derivative of cosine of x equal to sine of x instead of writing cosine of x is equal to negative of sine of x. We must take care of these negative signs as their ignorance lead us to incorrect solutions.