Question

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

Answer 1

Hint: 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 $ .