Solving a Tricky Integration Problem: A Comprehensive Guide
Today, we tackle a complex integration problem that requires some clever manipulation and substitution. If you're looking to deepen your understanding of calculus, particularly integration involving trigonometric functions, this article is for you!
The Problem and the Solution
Consider the problem:
Find the integral (I int frac{sec^2 x}{sec x tan^{9/5} x} , dx).
This problem can be approached by making a series of substitutions, which requires persistence and a keen eye for patterns.
Step-by-Step Solution
The first step is to make a substitution that simplifies the integral. Let's start with:
(sec x tan x z)
By differentiating both sides, we get:
[(sec x tan x)sec^2 x , dx dz]
Therefore, we can express (sec x , dx) as:
(sec x , dx frac{dz}{z})
Substituting this into the original integral, we get:
(I int frac{sec x}{z cdot z^{9/5}} , dz int frac{1}{z cdot z^{9/5}} , dz int z^{-14/5} , dz)
This integral can be solved by using the power rule for integration:
(I frac{z^{-9/5}}{-9/5} C -frac{5}{9z^{9/5}} C)
Alternative Approach
An alternative approach is to use the known identity:
(sec^2 x - tan^2 x 1)
Thus, we can write:
[sec x - tan x frac{1}{z}]
Adding this to the substitution, we get:
[2sec x frac{z^2 1}{z}]
This allows us to express:
[sec x frac{1}{2}left(frac{z^2 1}{z}right)]
Substituting back into the integral:
[I frac{1}{2} int frac{z^2 1}{z^{13/2}} , dz frac{1}{2} left[frac{z^{-3/2}}{-3/2} - frac{z^{-5/2}}{-5/2}right] C -frac{1}{3z^{-3/2}} - frac{1}{5z^{-5/2}} C]
Finally, substituting back (z sec x tan x) gives us the final answer:
[I -frac{1}{7 sec x tan^{7/2} x} - frac{1}{11 sec x tan^{11/2} x} C]
Conclusion
This integration problem required careful manipulation and multiple substitutions, highlighting the importance of recognizing patterns and utilizing identities. If you found this helpful, you might also want to explore similar problems or delve deeper into integration techniques.