sinx 6. S dx cos² x

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please show how to integrate by hand
### Problem 6

Evaluate the integral:

\[ \int \frac{\sin x}{\cos^2 x} \, dx \]

#### Explanation:

This problem involves finding the indefinite integral of the given function. The integrand is \(\frac{\sin x}{\cos^2 x}\), which is a rational function involving trigonometric functions. 

An applicable technique for solving this integral is to use a substitution method, particularly recognizing a derivative of a basic trigonometric function or manipulating the integrand into a familiar form. 

For instance, let \( u = \cos x \). Then \( du = -\sin x \, dx \), and \(\sin x \, dx = -du\).

Rewriting the integral in terms of \( u \):

\[
\int \frac{\sin x}{\cos^2 x} \, dx = \int \frac{1}{u^2} (-du) = -\int u^{-2} \, du
\]

Using the power rule for integration:

\[
-\int u^{-2} \, du = - \left( \frac{u^{-1}}{-1} \right) + C = \frac{1}{u} + C
\]

Substituting back \( u = \cos x \):

\[
\int \frac{\sin x}{\cos^2 x} \, dx = \frac{1}{\cos x} + C = \sec x + C
\]

Thus, the solution to the integral is:

\[
\int \frac{\sin x}{\cos^2 x} \, dx = \sec x + C
\]
Transcribed Image Text:### Problem 6 Evaluate the integral: \[ \int \frac{\sin x}{\cos^2 x} \, dx \] #### Explanation: This problem involves finding the indefinite integral of the given function. The integrand is \(\frac{\sin x}{\cos^2 x}\), which is a rational function involving trigonometric functions. An applicable technique for solving this integral is to use a substitution method, particularly recognizing a derivative of a basic trigonometric function or manipulating the integrand into a familiar form. For instance, let \( u = \cos x \). Then \( du = -\sin x \, dx \), and \(\sin x \, dx = -du\). Rewriting the integral in terms of \( u \): \[ \int \frac{\sin x}{\cos^2 x} \, dx = \int \frac{1}{u^2} (-du) = -\int u^{-2} \, du \] Using the power rule for integration: \[ -\int u^{-2} \, du = - \left( \frac{u^{-1}}{-1} \right) + C = \frac{1}{u} + C \] Substituting back \( u = \cos x \): \[ \int \frac{\sin x}{\cos^2 x} \, dx = \frac{1}{\cos x} + C = \sec x + C \] Thus, the solution to the integral is: \[ \int \frac{\sin x}{\cos^2 x} \, dx = \sec x + C \]
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