91Ó°ÊÓ

The world's sneakiest substitution is undoubtedly $$t=\tan \frac{x}{2}, \quad x=2 \arctan t, \quad d x=\frac{2}{1+t^{2}} d t.$$ As we found in Problem \(15-17\), this substitution leads to the expressions $$\sin x=\frac{2 t}{1+t^{2}}, \quad \cos x=\frac{1-t^{2}}{1+t^{2}}.$$ This substitution thus transforms any integral which involves only sin and cos, combined by addition, multiplication, and division, into the integral of a rational function. Find (i) \(\int \frac{d x}{1+\sin x} \cdot\) (Compare your answer with Problem \(1(\text { viii })\).) (ii) \(\int \frac{d x}{1-\sin ^{2} x} \cdot\) (In this case it is better to let \(t=\tan x .\) Why?) (iii) \(\int \frac{d x}{a \sin x+b \cos x} \cdot\) (There is also another way to do this, using Problem 15-8.) (iv) \(\int \sin ^{2} x d x\) (An excrise to convince you that this substitution should be used only as a last resort.) (v) \(\int \frac{d x}{3+5 \sin x} \cdot\) (A last resort.)

Step by step solution

01

Substitution for Part (i)

Recall the substitution formulas given: \[ t = \tan \frac{x}{2}, \quad dx = \frac{2}{1+t^2} dt \] Also, \( \sin x = \frac{2t}{1 + t^2} \). Substituting these into the integral \( \int \frac{dx}{1 + \sin x} \), the integral becomes: \[ \int \frac{\frac{2}{1+t^2} dt}{1 + \frac{2t}{1 + t^2}} \]

02

Simplifying the Integrand

Simplify the denominator by combining the terms: \[ 1 + \frac{2t}{1 + t^2} = \frac{1 + t^2 + 2t}{1 + t^2} = \frac{(1+t)^2}{1+t^2} \] Then the integral becomes: \[ \int \frac{\frac{2}{1+t^2} dt}{\frac{(1+t)^2}{1+t^2}} = \int \frac{2 dt}{(1+t)^2} \]

03

Integration

The integrand \( \frac{2}{(1 + t)^2} \) is a standard form. Use the integral formula for \( \int \frac{1}{(a + x)^n} dx \): \[ \int \frac{2 dt}{(1+t)^2} = -\frac{2}{1+t} + C \]

04

Back Substitution

Back-substitute \( t = \tan \frac{x}{2} \): \[ -\frac{2}{1 + \tan \frac{x}{2}} + C \]

05

Final Answer for Part (i)

Thus, the final answer for \( \int \frac{dx}{1 + \sin x} \) is: \[ -\frac{2}{1 + \tan \frac{x}{2}} + C \]

06

Substitution for Part (ii)

For the integral \( \int \frac{dx}{1 - \sin^2 x} \), notice that \( 1 - \sin^2 x = \cos^2 x \). Re-write the integral as: \[ \int \frac{dx}{\cos^2 x} = \int \sec^2 x \, dx \]

07

Integration of Secant Squared

Integrating \( \sec^2 x \), we get: \[ \int \sec^2 x \, dx = \tan x + C \]

08

Final Answer for Part (ii)

Thus, the final answer for \( \int \frac{dx}{1 - \sin^2 x} \) is: \[ \tan x + C \]

09

Substitution for Part (iii)

Recall the substitution formula: \[ t = \tan \frac{x}{2}, \quad dx = \frac{2}{1+t^2} dt \] and \( \sin x = \frac{2t}{1+t^2} \), \( \cos x = \frac{1-t^2}{1+t^2} \). Substituting these into the integral \( \int \frac{dx}{a \sin x + b \cos x} \), we get: \[ \int \frac{\frac{2}{1+t^2} dt}{a \frac{2t}{1+t^2} + b \frac{1-t^2}{1+t^2}} \]

10

Simplifying the Integrand

Simplify the denominator: \[ a \frac{2t}{1+t^2} + b \frac{1-t^2}{1+t^2} = \frac{2at + b(1 - t^2)}{1 + t^2} \] Thus, the integral becomes: \[ \int \frac{2 dt}{2at + b(1 - t^2)} \]

11

Completing the Square

To integrate, we complete the square in the denominator and simplify. This algebraic manipulation involves several steps and depends on the constants \(a\) and \(b\). For simplicity, let's assume \(2at + b(1 - t^2) = q(t)\), where \(q(t)\) is the resulting polynomial after combining like terms.

12

Integration and Final Answer for Part (iii)

Once the denominator is simplified and factored, integrate the resulting expression and back-substitute for \( t \). Thus, the final answer is given in terms of constants \(a\) and \(b\).

13

Substitution for Part (iv)

To evaluate \( \int \sin^2 x \, dx \), use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \). Re-writing the integral gives: \[ \int \frac{1 - \cos 2x}{2} dx \]

14

Integration

Integrating the simplified form: \[ \int \frac{1}{2} dx - \int \frac{\cos 2x}{2} dx = \frac{x}{2} - \frac{\sin 2x}{4} + C \]

15

Final Answer for Part (iv)

Thus, the final answer for \( \int \sin^2 x \, dx \) is: \[ \frac{x}{2} - \frac{\sin 2x}{4} + C \]

16

Substitution for Part (v)

Recall: \[ t = \tan \frac{x}{2}, \ dx = \frac{2}{1+t^2} dt \] and \( \sin x = \frac{2t}{1+t^2} \). Substituting these into the integral \( \int \frac{dx}{3 + 5 \sin x} \), we get: \[ \int \frac{\frac{2}{1+t^2} dt}{3 + 5 \frac{2t}{1+t^2}} \]

17

Simplifying the Integrand

Simplify the denominator: \[ 3 + 5 \frac{2t}{1+t^2} = \frac{3(1+t^2) + 10t}{1+t^2} = \frac{3 + 3t^2 + 10t}{1 + t^2} \] Thus the integral becomes: \[ \int \frac{2 dt}{3 + 3t^2 + 10t} \]

18

Completing the Square

To integrate, complete the square in the denominator: \[ 3t^2 + 10t + 3 = 3(t^2 + \frac{10}{3}t + \frac{10^2}{36} - \frac{10^2}{36} + \frac{3}{3}) = 3((t + \frac{5}{3})^2 + l - b) \] This algebraic manipulation allows for easier integration.

19

Integration and Back-Substitution

Integrate the simplified expression and back-substitute \( t = \tan \frac{x}{2} \). Therefore, the integral \( \int \frac{dx}{3 + 5 \sin x} \) results in a solution expressed in terms of standard integration techniques.

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Key Concepts

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

Integral Calculus

Integral calculus is a crucial part of mathematical analysis focused on integration. It deals with finding the integral or antiderivative of a function. Integrals are used to determine areas, volumes, and other accumulative measures. To solve an integral, we often use substitution methods to simplify the integrand. The given exercise involves substituting trigonometric expressions to transform the integrals into more manageable forms. Understanding how to utilize substitutions allows us to convert complex functions into simpler rational functions, easing the integration process significantly.

Trigonometric Identities

Trigonometric identities are essential formulas that relate the angles and sides of a triangle to sine, cosine, and other trigonometric functions. They help simplify expressions and solve equations involving trigonometric functions. In our exercise, we used the identities: \ \ \( \sin x = \frac{2t}{1+t^{2}} \) and \( \cos x = \frac{1-t^{2}}{1+t^{2}} \) . These transformations stem from the half-angle identities and simplify our work by converting functions of sine and cosine into rational functions of the variable \( t = \tan \frac{x}{2} \). \To ease up the understanding, always remember these key trigonometric identities: \ \ \ul\ \