
What is the antiderivative of $\dfrac{\sin x}{1+\cos x}$ ?
Answer
480.9k+ views
Hint: Here we have been given a value and we have to find its antiderivative. Anti-derivative means we have to integrate the given value. So we will use a substitution method for integrating the given value. Firstly we will let the denominator equal to some variable and differentiate both sides we respect to $x$ . Then we will replace the value obtained inside the integration and solve it. Finally we will put the left value back in the answer and get our final answer.
Complete step-by-step solution:
We have to find the antiderivative of below value:
$\dfrac{\sin x}{1+\cos x}$
So we have to find,
$\int{\dfrac{\sin x}{1+\cos x}dx}$….$\left( 1 \right)$
We will use a substitution method to solve the above integral.
Firstly let,
$1+\cos x=u$…$\left( 2 \right)$
Differentiate both sides with respect to $x$ :
$\dfrac{d}{dx}\left( 1+\cos x \right)=\dfrac{d}{dx}u$
We know the differentiation of constant is zero and differentiation of cosine is minus sine.
$\Rightarrow 0-\sin x=\dfrac{du}{dx}$
$\Rightarrow -\sin x=\dfrac{du}{dx}$
Take the denominator value from the right side to the left side,
$\Rightarrow -\sin x\,dx=du$
Multiplying both side by $-1$ we get,
$\Rightarrow \sin x\,dx=-du$….$\left( 3 \right)$
Substitute the value from equation (2) and (3) in equation (1) we get,
$\Rightarrow \int{\dfrac{-du}{u}}$
We know$\int{\dfrac{1}{x}dx=\log x+c}$ where $c$ is constant using it above we get,
$\Rightarrow -\log u+c$
Where $c$ is any constant
Substitute the value from equation (1) above we get,
$\Rightarrow -\log \left( 1+\cos x \right)+c$
So we get the answer as $-\log \left( 1+\cos x \right)+c$
Hence the antiderivative of $\dfrac{\sin x}{1+\cos x}$ is $-\log \left( 1+\cos x \right)+c$ where $c$ is any constant.
Note: Anti derivatives are the inverse derivatives or indefinite integral of a function. Anti derivatives can be used to find the definite integral using the fundamental theorem of calculus. We can also rationalize the denominator method in this question and use the various formulas of trigonometric functions to get our answer.
Complete step-by-step solution:
We have to find the antiderivative of below value:
$\dfrac{\sin x}{1+\cos x}$
So we have to find,
$\int{\dfrac{\sin x}{1+\cos x}dx}$….$\left( 1 \right)$
We will use a substitution method to solve the above integral.
Firstly let,
$1+\cos x=u$…$\left( 2 \right)$
Differentiate both sides with respect to $x$ :
$\dfrac{d}{dx}\left( 1+\cos x \right)=\dfrac{d}{dx}u$
We know the differentiation of constant is zero and differentiation of cosine is minus sine.
$\Rightarrow 0-\sin x=\dfrac{du}{dx}$
$\Rightarrow -\sin x=\dfrac{du}{dx}$
Take the denominator value from the right side to the left side,
$\Rightarrow -\sin x\,dx=du$
Multiplying both side by $-1$ we get,
$\Rightarrow \sin x\,dx=-du$….$\left( 3 \right)$
Substitute the value from equation (2) and (3) in equation (1) we get,
$\Rightarrow \int{\dfrac{-du}{u}}$
We know$\int{\dfrac{1}{x}dx=\log x+c}$ where $c$ is constant using it above we get,
$\Rightarrow -\log u+c$
Where $c$ is any constant
Substitute the value from equation (1) above we get,
$\Rightarrow -\log \left( 1+\cos x \right)+c$
So we get the answer as $-\log \left( 1+\cos x \right)+c$
Hence the antiderivative of $\dfrac{\sin x}{1+\cos x}$ is $-\log \left( 1+\cos x \right)+c$ where $c$ is any constant.
Note: Anti derivatives are the inverse derivatives or indefinite integral of a function. Anti derivatives can be used to find the definite integral using the fundamental theorem of calculus. We can also rationalize the denominator method in this question and use the various formulas of trigonometric functions to get our answer.
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