Question

How do you find the integral of \[\int {{{\sec }^2}\left( {5x} \right)} dx\]?

Answer 1

Hint: Here we will integrate the trigonometric function by using the substitution method. We will assume the angle of the given trigonometric function to be any variable and then we will find its derivative. We will then substitute the angle in terms of the variable and its derivative in the given function and solve it using the integration formula to get the required value of the given integration.

Complete step by step solution:
Here we need to find the value of the integration of the trigonometric function.
We need to find the value of \[\int {{{\sec }^2}\left( {5x} \right)} dx\].
Let the value of \[\int {{{\sec }^2}\left( {5x} \right)} dx\] be \[I\].
We can write it as
\[I = \int {{{\sec }^2}\left( {5x} \right)} dx\] …………. \[\left( 1 \right)\]
Here, we will use the substitution method of integration.
Let \[5x = t\]
Now, we will differentiate both sides.
\[ \Rightarrow 5 \times dx = dt\]
Now, we will divide both sides by 5.
\[\begin{array}{l} \Rightarrow \dfrac{{5 \times dx}}{5} = \dfrac{{dt}}{5}\\ \Rightarrow dx = \dfrac{{dt}}{5}\end{array}\]
Now, we will substitute \[5x = t\] and \[dx = \dfrac{{dt}}{5}\] in the equation \[\left( 1 \right)\]. Therefore, we get
\[I = \int {{{\sec }^2}t} \times \dfrac{{dt}}{5}\]
Now, we will take the constant term outside from the integration.
\[ \Rightarrow I = \dfrac{1}{5}\int {{{\sec }^2}t} dt\]
Integrating the trigonometric function using the formula \[\int {{{\sec }^2}\theta d\theta } = \tan \theta \], we get
\[ \Rightarrow I = \dfrac{1}{5} \times \tan t + c\]
where c is the integral constant. Now, we will substitute the value of \[t\] here.
\[ \Rightarrow I = \dfrac{{\tan \left( {5x} \right)}}{5} + c\]
Hence, this is the required value of given integrals.

Note:
Here we have obtained the value of the given integration. The integration represents the summation of discrete data. The integral is used to find the functions which will describe the displacement, area, volume, which occurs due to a collection of small data and cannot be measured singularly. There are basically two types of integrals and they are Definite integrals and indefinite integrals. The integration is an inverse of differentiation and hence it is called antiderivative.