Proof of Integral of sinx formula

$x$ is a variable, which is considered as an angle of a right triangle and the sine function is written as $\sin{x}$ in trigonometric mathematics. The indefinite integral of $\sin{x}$ with respect to $x$ is written as follows to find the integration of sine function in calculus.

$\displaystyle \int{\sin{x} \,}dx$

Derivative of cos function

Write the derivative of cos function with respect to $x$ formula for expressing the differentiation of cosine function in mathematical form.

$\dfrac{d}{dx}{\, \cos{x}} \,=\, -\sin{x}$

$\implies$ $\dfrac{d}{dx}{(-\cos{x})} \,=\, \sin{x}$

Inclusion of an Arbitrary constant

According to differential calculus, the derivative of a constant is always zero. So, it doesn’t affect the process of the differentiation if an arbitrary constant $(c)$ is added to the trigonometric function $-\cos{x}$.

$\implies$ $\dfrac{d}{dx}{(-\cos{x}+c)} \,=\, \sin{x}$

Integral of sin function

The collection of all primitives of $\sin{x}$ function is called the indefinite integral of $\sin{x}$ function, which is written in the following mathematical form in integral calculus.

$\displaystyle \int{\sin{x} \,}dx$

In this case, the primitive or an antiderivative of $\sin{x}$ is $-\cos{x}$ and the constant of integration $c$.

$\dfrac{d}{dx}{(-\cos{x}+c)} = \sin{x}$ $\,\Longleftrightarrow\,$ $\displaystyle \int{\sin{x} \,}dx = -\cos{x}+c$

$\therefore \,\,\,\,\,\,$ $\displaystyle \int{\sin{x} \,}dx = -\cos{x}+c$

Therefore, it is proved that the antiderivative or indefinite integration of sine function is equal to the sum of the negative cos function and the constant of integration.

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