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Question: How do you integrate \(\int {{{\sec }^2}\left( {2x - 1} \right)} dx\) ?...

How do you integrate ∫sec⁡2(2x−1)dx\int {{{\sec }^2}\left( {2x - 1} \right)} dx ?

Explanation

Solution

In this question, we are given an expression and we have been asked to integrate it. But, direct integration of the given expression does not exist. So, we will have to use some of the tricks to find the answer. Use substitution on the angle of the trigonometric expression to simplify it. Assume it equal to a variable and then, differentiate both the sides. Put the values in the given expression and then, integrate. You will have your answer.

Formula used: ∫sec⁡2xdx=tan⁡x+C\int {{{\sec }^2}xdx = \tan x + C}

Complete step-by-step solution:
We are given a trigonometric expression and we have to find its integration.
We know that a direct integration of sec⁡2x{\sec ^2}x exists. But what about 2x−12x - 1 as its angle
We will put it equal to a variable in order to simplify the equation.
⇒∫sec⁡2(2x−1).dx\Rightarrow \int {{{\sec }^2}\left( {2x - 1} \right)} .dx …. (given)
Let 2x−1=v2x - 1 = v ………..…. (1)
Differentiating both the sides with respect to xx,
⇒ddx(2x−1)=dvdx\Rightarrow \dfrac{d}{{dx}}\left( {2x - 1} \right) = \dfrac{{dv}}{{dx}}
We will get,
⇒2=dvdx\Rightarrow 2 = \dfrac{{dv}}{{dx}}
Shifting the denominator to the other side in order to find the value of dx,
⇒dx=dv2\Rightarrow dx = \dfrac{{dv}}{2} ………..…. (2)
Now we will put equation (1) and equation (2) in the given expression,
⇒12∫sec⁡2vdv\Rightarrow \dfrac{1}{2}\int {{{\sec }^2}v} dv
Now, we have direct integration of sec⁡2x{\sec ^2}x. Using that, we will get –
⇒tan⁡v2+C\Rightarrow \dfrac{{\tan v}}{2} + C …. (∫sec⁡2xdx=tan⁡x+C)\left( {\int {{{\sec }^2}xdx = \tan x + C} } \right)
Putting the value of v (2x−1=v2x - 1 = v),
⇒tan⁡(2x−1)2+C\Rightarrow \dfrac{{\tan \left( {2x - 1} \right)}}{2} + C

Therefore, the value of the given integral is tan⁡(2x−1)2+C \dfrac{{\tan \left( {2x - 1} \right)}}{2} + C

Note: We are told to add a constant of integration at the end of every answer. But why so,
The answer to this lies in differentiation. In order to understand the logic behind this, you also need to know that integration is also called anti-derivative.
Now, let us see an example. Differentiation of x3+44{x^3} + 44 is 3x23{x^2} and similarly, differentiation of x3+6{x^3} + 6 is also 3x23{x^2}. If we integrate 3x23{x^2}, we will get x3{x^3} but how will we know whether it had +44 + 44 or +6 + 6 at the end? For this purpose, we add C (constant of integration) at the end.