/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Evaluate the indefinite integral... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the indefinite integral. \(\int \frac{d t}{\cos ^{2} t \sqrt{1+\tan t}}\)

Short Answer

Expert verified
The indefinite integral is \(2\sqrt{1+\tan t} + C\).

Step by step solution

01

Substitute for Simplification

Let \( u = \tan t \). Then we know that \( du = \sec^2 t \, dt = \frac{dt}{\cos^2 t} \). Therefore, \( dt = \cos^2 t \, du \). Substitute into the integral to get: \[ \int \frac{dt}{\cos^2 t \sqrt{1+\tan t}} = \int \frac{\cos^2 t}{\cos^2 t \sqrt{1+u}} \, du = \int \frac{1}{\sqrt{1+u}} \, du. \]
02

Integral Simplification

The integral now becomes \( \int \frac{1}{\sqrt{1+u}} \, du \). This is a standard integral that can be solved by substitution or by recognizing it as a form of the integral for inverse hyperbolic functions.
03

Solve the Integral

The antiderivative of \( \frac{1}{\sqrt{1+u}} \) is \( 2\sqrt{1+u} + C \), where \( C \) is the constant of integration. Thus, \( \int \frac{1}{\sqrt{1+u}} \, du = 2\sqrt{1+u} + C \).
04

Substitute Back

Replace \( u = \tan t \) back into our solution to get \( 2\sqrt{1+\tan t} + C \). Thus, the integral evaluates to \( 2\sqrt{1+\tan t} + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Substitution
Trigonometric substitution is a powerful technique used in calculus to simplify integrals involving trigonometric functions. It involves substituting a trigonometric identity to transform a complex expression into a simpler form. This method can be particularly useful when dealing with integrals that involve radicals, trigonometric powers, or both, as seen in our exercise.

In this problem, we started with the integral \( \int \frac{d t}{\cos ^2 t \sqrt{1+\tan t}} \). By letting \( u = \tan t \), we used the identity \( dt = \cos^2 t \, du \) to simplify the expression within the integral. This transformation switches the variable of integration from \( t \) to \( u \), allowing us to eliminate the trigonometric functions from the integral, leading to a more manageable form: \( \int \frac{1}{\sqrt{1+u}} \, du \).

This kind of substitution is beneficial because it often turns a difficult integral into one that can be easily identified and computed using standard techniques and tables.
Calculus Problem-Solving
When it comes to calculus problem-solving, understanding the fundamentals of integration and differential calculus is crucial. It's not just about solving equations; it's about knowing when to apply specific techniques to simplify and evaluate problems.

In our exercise, we demonstrated problem-solving by strategically choosing the right substitution to simplify the integral. This involves a deep understanding of function behavior, trigonometric identities, and the manipulation of algebraic expressions. The choice of substituting \( u = \tan t \) was not arbitrary—it was aimed at simplifying the original complex expression into a form that is more straightforward to integrate.

Developing these problem-solving skills requires practice and an ability to recognize patterns and forms that appear in many problems, like inverse trigonometric and hyperbolic identities. Through repetitive solving and understanding the reasons behind each transformation, you can gain confidence and proficiency in tackling challenging integrals.
Integration Techniques
Integration techniques are varied methods used to compute integrals. Some common techniques include substitution, integration by parts, partial fraction decomposition, and trigonometric identities, among others. Choosing the right technique is essential to efficiently solve an integral.

In the exercise, we utilized the technique of trigonometric substitution. After reducing the expression with \( u = \tan t \), we faced the integral \( \int \frac{1}{\sqrt{1+u}} \, du \). Solving this required recognizing it as a standard integral related to inverse hyperbolic functions. The solution was derived as \( 2\sqrt{1+u} + C \).

Key integration techniques include:
  • Substitution: Replaces a variable with another to simplify the integral.
  • Integration by parts: Breaks down products of functions into simpler components.
  • Partial fraction decomposition: Used primarily for rational functions to break them into simpler fractions.
  • Trigonometric identities: Applies known identities to simplify the integrand.
By mastering these techniques, one can solve a wide range of integrals encountered in calculus.

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