Integrate Tan/Sin Without Special Methods

  • Thread starter tandoorichicken
  • Start date
  • Tags
    Integration
In summary, integrating tan(x) can be done by using the substitution method and the trigonometric identity 1 + tan^2(x) = sec^2(x). Similarly, integrating sin(x)/cos(x) also involves using the substitution method and the trigonometric identity 1 + tan^2(x) = sec^2(x). However, integrating tan(x)/sin(x) and tan(x)*sin(x) requires special methods such as substitution or integration by parts. The easiest way to integrate tan(x)*sin(x) is to use the product-to-sum identity and then the substitution method and trigonometric identity 1 + tan^2(x) = sec^2(x). There is no specific rule for integrating tan(x
  • #1
tandoorichicken
245
0
How do you do the following integral without using any special methods, i.e., integration by parts, etc.

[tex]\int \frac{\tan x}{\sin x} \,dx[/tex]

?
 
Physics news on Phys.org
  • #2
What does [tex]\tan x[/tex] equal? Hint: [tex]\tan x = \frac{\sin x}{?}[/tex] You should remember what the "?" is! :tongue2:

Anyway, after you make that simplification you will have [tex]\int \sec x \ dx[/tex]

Unfortunately this integral is tricky to evaluate and DOES require some special tricks to solve (not integration by parts though). Check out this website to see how: integral of secant
 
Last edited:
  • #3


To integrate this integral without using any special methods, we can use the substitution method. Let u = sin x, then du = cos x dx. We can rewrite the integral as:

\int \frac{\tan x}{\sin x} \,dx = \int \frac{\tan x}{u} \,du

Next, we can use the trigonometric identity tan x = sin x / cos x to simplify the integrand:

\int \frac{\tan x}{u} \,du = \int \frac{\sin x}{u \cos x} \,du

Using the substitution u = sin x, we can also rewrite cos x as \sqrt{1-u^2}:

\int \frac{\sin x}{u \cos x} \,du = \int \frac{1}{u \sqrt{1-u^2}} \,du

Now, we can use the substitution v = 1-u^2, then dv = -2u du:

\int \frac{1}{u \sqrt{1-u^2}} \,du = -\frac{1}{2} \int \frac{1}{\sqrt{v}} \,dv

Solving this integral, we get:

-\frac{1}{2} \int \frac{1}{\sqrt{v}} \,dv = -\frac{1}{2} \sqrt{v} + C = -\frac{1}{2} \sqrt{1-u^2} + C

Substituting back u = sin x, we get the final answer of:

\int \frac{\tan x}{\sin x} \,dx = -\frac{1}{2} \sqrt{1-\sin^2 x} + C = -\frac{1}{2} \sqrt{1-\frac{\sin^2 x}{1}} + C = -\frac{1}{2} \sqrt{1-\cos^2 x} + C = -\frac{1}{2} \sin x + C

Therefore, we have successfully integrated \int \frac{\tan x}{\sin x} \,dx without using any special methods. This approach may require some extra steps and algebraic manipulation, but it is a useful method to integrate trigonometric functions without relying on integration by parts or other special methods.
 

1. How do you integrate tan(x)?

Integrating tangent (tan) functions can be done by using the substitution method, where you substitute u = tan(x) and then use the trigonometric identity 1 + tan^2(x) = sec^2(x) to simplify the integral.

2. What is the formula for integrating sin(x)/cos(x)?

The formula for integrating sine (sin) over cosine (cos) is to use the substitution method, where you substitute u = cos(x) and then use the trigonometric identity 1 + tan^2(x) = sec^2(x) to simplify the integral.

3. Can you integrate tan(x)/sin(x) without using special methods?

No, you cannot integrate tangent over sine without using special methods such as substitution or integration by parts. This is because the integral involves a quotient of trigonometric functions, which cannot be solved using basic integration rules.

4. What is the easiest way to integrate tan(x)*sin(x)?

The easiest way to integrate tangent (tan) multiplied by sine (sin) is to use the product-to-sum identity: tan(x)*sin(x) = (1/2)[sin(2x)/cos(x)]. Then, you can integrate using the substitution method and the trigonometric identity 1 + tan^2(x) = sec^2(x).

5. Is there a specific rule for integrating tan(x)*sin(x)?

There is no specific rule for integrating tangent (tan) multiplied by sine (sin). However, you can use the product-to-sum identity and then solve the integral using the substitution method and trigonometric identities.

Similar threads

Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
28
Views
263
  • Introductory Physics Homework Help
Replies
4
Views
166
  • Calculus and Beyond Homework Help
Replies
3
Views
269
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
406
  • Introductory Physics Homework Help
Replies
10
Views
112
  • Introductory Physics Homework Help
Replies
8
Views
716
Back
Top