Question

# Evaluate the given integral: $\int {{{\sec }^2}(7 - 4x)dx}$

Hint: In this question we have to find the integral of the given quantity. Consider the given quantity is equal to some variable and then use the substitution method, i.e., put $7 - 4x = t$. Then integrate and substitute back to get an answer in ‘$x$’.

Consider the integral given as $I$
$\Rightarrow I = \int {{{\sec }^2}(7 - 4x)dx}$---(1)
Now put $7 - 4x = t$---(2)
On differentiating both sides, we get,
$\Rightarrow - 4dx = dt$
$\Rightarrow dx = - \dfrac{1}{4}dt$
After substituting $(7 - 4x)$ as ‘$t$’ and $dx = - \dfrac{1}{4}dt$ in equation (1) we get,
$\Rightarrow I = - \dfrac{1}{4}\int {{{\sec }^2}t{\kern 1pt} dt}$
We know that, $\int {{{\sec }^2}xdx = \tan x + c}$
$\Rightarrow I = - \dfrac{1}{4}\tan t + c$
From equation (2) we get,
$\Rightarrow I = - \dfrac{1}{4}\tan (7 - 4x) + c$
Where, $c$ is constant
Thus, $\int {{{\sec }^2}(7 - 4x)dx = - \dfrac{1}{4}\tan (7 - 4x) + c}$

Note: Whenever you face such types of problems the key concept is simply to make use of the substitution method. And it is important to add constant after integrating the indefinite integral. Do not forget to differentiate after substitution to find ‘$dx$’.