Evaluate the following integral.
{eq}\displaystyle \int \dfrac { \cot x \ dx }{ \csc x + 16 \sin x} {/eq}
Question:
Evaluate the following integral.
{eq}\displaystyle \int \dfrac { \cot x \ dx }{ \csc x + 16 \sin x} {/eq}
Anti-Derivative:
The anti-derivative is an operation that can be performed on many functions. An anti-derivative gets a function whose derivative is the function whose anti-derivative is taken. In other words, the anti-derivative is an opposite process to differentiation or taking the derivative.
Answer and Explanation: 1
Given:
{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx {/eq}
Expand the cotangent term. Recall:
{eq}\displaystyle \cot\ x = \frac{\cos\ x}{\sin\ x} {/eq}
{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \int\ \frac{\cos\ x}{\sin\ x(\csc\ x + 16\sin\ x)}\ dx {/eq}
{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \int\ \frac{\cos\ x}{\sin\ x\csc\ x + 16\sin^2\ x}\ dx {/eq}
Recall that:
{eq}\displaystyle \sin\ x\csc\ x = 1 {/eq}
Therefore,
{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \int\ \frac{\cos\ x}{1 + 16\sin^2\ x}\ dx {/eq}
Let:
{eq}\displaystyle u = 4\sin\ x {/eq}
{eq}\displaystyle du = 4\cos\ x\ dx {/eq}
Or:
{eq}\displaystyle \frac{1}{4}\ du = \cos\ x\ dx {/eq}
In terms of u, we can write this integral as:
{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \frac{1}{4}\int\ \frac{1}{1+ u^2}\ du {/eq}
This is a standard integral:
{eq}\displaystyle \int\ \frac{1}{1+ u^2}\ du = \tan^{-1}\ u + C {/eq}
Therefore,
{eq}\displaystyle \int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \frac{1}{4}\tan^{-1}\ u + C {/eq}
Return the definition of u to get:
{eq}\displaystyle \boxed{\int\ \frac{\cot\ x}{\csc\ x + 16\sin\ x}\ dx = \frac{1}{4}\tan^{-1}\ (4\sin\ x) + C} {/eq}
Learn more about this topic:
from
Chapter 13 / Lesson 13Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.