Math Problem Statement

Solve the following integration

Solve the integral \( \int \frac{dx}{1 + \sin x} \).

Solution

The integral provided in the image is:

∫dx1+sin⁡x\int \frac{dx}{1 + \sin x}

Solution:

We will simplify and solve this integral step by step:

Step 1: Simplify the denominator 1+sin⁡x1 + \sin x:

Use the trigonometric identity: 1+sin⁡x=1+2sin⁡x2cos⁡x2=(cos⁡2x2+sin⁡2x2)+2sin⁡x2cos⁡x21 + \sin x = 1 + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \left( \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \right) + 2 \sin \frac{x}{2} \cos \frac{x}{2}

This simplifies to: 1+sin⁡x=(cos⁡x2+sin⁡x2)21 + \sin x = \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2

Step 2: Substitute:

Let: t=tan⁡x2,so thatdx=21+t2 dtandsin⁡x=2t1+t2t = \tan \frac{x}{2}, \quad \text{so that} \quad dx = \frac{2}{1+t^2} \, dt \quad \text{and} \quad \sin x = \frac{2t}{1+t^2}

From this substitution, 1+sin⁡x1 + \sin x becomes: 1+sin⁡x=1+2t1+t2=1+t2+2t1+t2=(1+t)21+t21 + \sin x = 1 + \frac{2t}{1+t^2} = \frac{1+t^2 + 2t}{1+t^2} = \frac{(1+t)^2}{1+t^2}

Thus: dx1+sin⁡x=2 dt1+t2(1+t)21+t2=2 dt(1+t)2\frac{dx}{1 + \sin x} = \frac{\frac{2 \, dt}{1+t^2}}{\frac{(1+t)^2}{1+t^2}} = \frac{2 \, dt}{(1+t)^2}

Step 3: Integrate:

Now we need to solve: ∫2 dt(1+t)2\int \frac{2 \, dt}{(1+t)^2}

Use the standard formula: \int \frac{1}{(1+t)^n} dt = \frac{(1+t)^{1-n}}{1-n} \quad \text{(for n \neq 1)}.

Here, n=2n = 2: ∫2 dt(1+t)2=−21+t+C\int \frac{2 \, dt}{(1+t)^2} = -\frac{2}{1+t} + C

Step 4: Back-substitute t=tan⁡x2t = \tan \frac{x}{2}:

= \dots Let’s complete the solution from where we left off in **Step 4**. We need to back-substitute $$t = \tan \frac{x}{2}$$ into the final result. Recall the integration result: \[ \int \frac{2 \, dt}{(1+t)^2} = -\frac{2}{1+t} + C.$$ --- ### Step 5: Back-substitution Substitute $$t = \tan \frac{x}{2}$$ back into the result: $$-\frac{2}{1+t} + C = -\frac{2}{1 + \tan \frac{x}{2}} + C.$$ Now, use the relation for $$1 + \tan \frac{x}{2}$$ in terms of trigonometric functions: $$1 + \tan \frac{x}{2} = \frac{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos^2 \frac{x}{2}} = \frac{(1 + \sin x)}{\cos^2 \frac{x}{2}}.$$ This simplifies the denominator: $$-\frac{2}{1 + \tan \frac{x}{2}} = -\frac{2 \cos^2 \frac{x}{2}}{1 + \sin x}.$$ --- ### Final Answer: The integral simplifies to: $$\int \frac{dx}{1 + \sin x} = -\frac{2 \cos^2 \frac{x}{2}}{1 + \sin x} + C.$$ Alternatively, this can be written using substitution identities depending on the context. --- Would you like me to expand on any specific step or reformat the result? Let me know! 😊 --- ### Related Questions: 1. How is the substitution $$t = \tan \frac{x}{2}$$ derived? 2. Can this integral be solved without substitution? If yes, how? 3. What are the practical applications of integrals involving trigonometric terms like $$1 + \sin x$$? 4. How does the identity $$1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$$ simplify solving? 5. How do definite integrals behave with such forms? --- ### Tip: When dealing with trigonometric integrals, substitutions like $$t = \tan \frac{x}{2}$$ often simplify the process dramatically by linearizing trigonometric functions!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integral calculus
Trigonometric identities
Substitution method

Formulas

\( 1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2 \)
Substitution: \( t = \tan \frac{x}{2}, \; dx = \frac{2}{1+t^2} dt \)
Integration formula: \( \int \frac{1}{(1+t)^n} dt = \frac{(1+t)^{1-n}}{1-n} \; (n \neq 1) \)

Theorems

Trigonometric substitution
Simplification using trigonometric identities

Suitable Grade Level

Undergraduate (Calculus I or II)

Related Recommendation

Solve the Integration \( \int \frac{dx}{1 + \sin x} \) Step-by-Step

Solve the Integral \( \int \frac{dx}{\sin{x} \cdot \sin{2x}} \) Using Trigonometric Substitution

Solve the Integral of sin(x)cos(x) / (sin(x) + cos(x))

Solve the Integral ∫ dx / (1 + sinx) Step-by-Step

Integral of 1 / (sin^2(x) cos^2(x)) Using Trigonometric Identities