Question

How do you integrate $\int {{{\sec }^2}(2x - 3)dx}$ using substitution?

Answer 1

First of all find the differentiation of the term given in the problem, $2x - 3$ with respect to $x$ and then use these values to substitute in the given integral and then find the integral after the substitution. Finally, re-substitute the value which was substituted in the obtained solution.

Complete step by step solution:
This will give the required result.
Complete step-by-step solution-
Consider the given integral as:
$I = \int {{{\sec }^2}(2x - 3)dx}$
The goal of the problem is to find the integral using the substitution method.
Therefore, let us consider $t = 2x - 3$
Differentiate both sides with respect to x.
$\dfrac{{dt}}{{dx}} = \dfrac{{d(2x - 3)}}{{dx}}$
$\dfrac{{dt}}{{dx}} = 2$
$\Rightarrow dx = \dfrac{1}{2}dt$
Now, substitute the above obtained result in the integral I, so we have
$I = \dfrac{1}{2}\int {{{\sec }^2}tdt}$
Integrate with respect to $t$ and obtain the integral.
Since, we know that $\dfrac{{d(\tan y)}}{{dy}} = {\sec ^2}y$
We get to the conclusion that
$I = \dfrac{1}{2}\tan t + c$ , where c is the constant of integration.
Now, substitute the value of $t$ into the equation:
$I = \dfrac{1}{2}\tan (2x - 3) + c$
The obtained integral of the given problem is $\dfrac{1}{2}\tan (2x - 3) + c$ , where $c$ is the integral constant
So, the correct answer is “ $\dfrac{1}{2}\tan (2x - 3) + c$ ”.

Note : The integration by substitution is also said as “The reverse chain rule”.
This is the method to integrate in some special cases. Let $f(g(x))$ be the integrand and we have to find the integral of the function $\left[ {f(g(x))g'(x)} \right]$ , then we use this method of integration.
So, the integral is given as:
$\Rightarrow \int {\left[ {f(g(x))g'(x)} \right]dx}$
Now, assume that $g(x) = t$ and differentiate both sides with respect to $x$ .