Question

How do you integrate $\int {\sec y(\tan y - \sec y)dy}$ ?

Answer 1

- In this question we have to evaluate the given integral, therefore, first we have to simplify the function by opening the bracket and performing basic mathematical operations and then take integral with respect to y on the simplified function.

Complete step by step solution:
Given integral
$I = \int {\sec y(\tan y - \sec y)dy}$
Firstly, let us open the bracket and simplify the function
$I = \int {(\sec y\tan y - {{\sec }^2}y)dy}$
Here, we can see that the function is of the form $f + g$ where $f$ and $g$ are two different functions.
Hence, we can apply the sum rule of integration which is given as follows:
$\int {f(x) \pm g(x)dx = \int {f(x)dx} \pm \int {g(x)dx} }$
On comparing we get
$f(y) = \sec y\tan y$
$g(y) = {\sec ^2}y$
Now let us integrate both functions one by one. So integrate $f(y)$ with respect to y we get
$\int {f(y)dy} = \int {\sec y\tan ydy}$
Since we know that $\dfrac{{d(\sec y)}}{{dy}} = \sec y\tan y$
We can write as
$\int {\sec y\tan ydy = \sec y + {c_{_1}}}$ ......(1)
where ${c_1}$ is the constant of integration.
Similarly, integrating $g(y)$ with respect to y we get
$\int {g(y)dy} = \int {{{\sec }^2}ydy}$
Since, we know that $\dfrac{{d(\tan y)}}{{dy}} = {\sec ^2}y$
We get to the conclusion that
$\int {{{\sec }^2}ydy = \tan y + {c_2}}$ ......(2)
On subtracting (1) and (2) we have
$\int {f(y)dy - \int {g(y)dy = \sec y - \tan y} } + {c_1} + {c_2}$
$\Rightarrow I = \sec y - \tan y + c$
Where $c = {c_1} + {c_2}$ and is the constant of integration.
Hence, this is our required answer.
So, the correct answer is “ $I = \sec y - \tan y + c$ ”.

Note : This is a common tendency among students to substitute $du$ or $dx$ while using the u-substitution method. However, this will lead you to a wrong answer. The operation of integration, up to an additive constant, is the inverse of the operation of differentiation.