$x$ is a variable and represents an angle of a right triangle. The secant and tangent functions are written as $\sec{x}$ and $\tan{x}$ respectively. The indefinite integration of product of secant and tan functions with respect to $x$ is written in integral calculus as follows.

$\displaystyle \int{\sec{x}\tan{x} \,}dx$

Now, let’s start deriving the proof for the indefinite integral of product of $\sec{x}$ and $\tan{x}$ functions with respect to $x$ in calculus.

As we know that the derivative of secant function is equal to the product of secant and tangent functions.

$\dfrac{d}{dx}{\, \sec{x}} \,=\, \sec{x}\tan{x}$

The derivative of a constant is zero. So, the derivative of secant function remains same even a constant is added to the differentiation of the secant function. Now, add an arbitrary constant to secant function and then differentiate it with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(\sec{x}+c)} \,=\, \sec{x}\tan{x}$

The collection of all primitives of product of $\sec{x}$ and $\tan{x}$ function is called as the integration of product of secant and tan functions. According to integral calculus, It can be expressed in mathematical form as follows.

$\displaystyle \int{\sec{x}\tan{x} \,}dx$

Therefore, the primitive or an antiderivative of $\sec{x}\tan{x}$ function is equal to the sum of the $\sec{x}$ function and the constant of integration ($c$).

$\dfrac{d}{dx}{(\sec{x}+c)} = \sec{x}\tan{x}$ $\,\Longleftrightarrow\,$ $\displaystyle \int{\sec{x}\tan{x} \,}dx = \sec{x}+c$

$\therefore \,\,\,\,\,\,$ $\displaystyle \int{\sec{x}\tan{x} \,}dx = \sec{x}+c$

Therefore, it has proved that the integration of product of secant and tan functions with respect to a variable is equal to the sum of the secant function and integral constant.

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