Integrate (1/(1+x^2) - cos(x)/sin^2(x)) dx
Understand the Problem
The question is asking to find the indefinite integral of the function (1/(1+x^2) - cos(x)/sin^2(x)). This involves recognizing standard integrals and applying appropriate integration techniques.
Answer
$\arctan(x) + \csc(x) + C$
Answer for screen readers
$\arctan(x) + \csc(x) + C$
Steps to Solve
- Split the integral
The integral of a difference is the difference of the integrals:
$$ \int \left( \frac{1}{1+x^2} - \frac{\cos(x)}{\sin^2(x)} \right) dx = \int \frac{1}{1+x^2} dx - \int \frac{\cos(x)}{\sin^2(x)} dx $$
- Evaluate the first integral
The integral of $\frac{1}{1+x^2}$ is a standard integral, which is the arctangent function:
$$ \int \frac{1}{1+x^2} dx = \arctan(x) + C_1 $$
- Evaluate the second integral
For the second integral, $\int \frac{\cos(x)}{\sin^2(x)} dx$, use a substitution. Let $u = \sin(x)$, so $du = \cos(x) dx$:
$$ \int \frac{\cos(x)}{\sin^2(x)} dx = \int \frac{1}{u^2} du = \int u^{-2} du $$
Now integrate with respect to $u$:
$$ \int u^{-2} du = -u^{-1} + C_2 = -\frac{1}{u} + C_2 $$
Substitute back $u = \sin(x)$:
$$ -\frac{1}{u} + C_2 = -\frac{1}{\sin(x)} + C_2 = -\csc(x) + C_2 $$
- Combine the results
Combine the results from the two integrals:
$$ \int \frac{1}{1+x^2} dx - \int \frac{\cos(x)}{\sin^2(x)} dx = \arctan(x) - (-\csc(x)) + C = \arctan(x) + \csc(x) + C $$
where $C = C_1 - C_2$ is the constant of integration.
$\arctan(x) + \csc(x) + C$
More Information
The indefinite integral represents a family of functions whose derivative is the original function. The "+ C" accounts for the arbitrary constant that disappears upon differentiation. The function $\csc(x)$ is the reciprocal of $\sin(x)$, i.e., $\csc(x) = \frac{1}{\sin(x)}$.
Tips
A common mistake is forgetting the constant of integration, $C$. Another common mistake is incorrectly evaluating the integral of $\frac{1}{1+x^2}$ or the integral involving trigonometric functions. Also, not recognizing the need for u-substitution in the second integral is a common error.
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