Integral Of Tan 2: Easy Solution Found

The quest to find the integral of tan(2x) has puzzled many a calculus student. It’s a problem that requires a combination of trigonometric identities and integration techniques. To tackle this, let’s first recall the double-angle formula for tangent:

tan(2x) = 2tan(x) / (1 - tan^2(x))

This formula can be rearranged to express tan(2x) in terms of tan(x), which will be useful for our integration.

Step 1: Express Tan(2x) in Terms of Tan(x)

Given the double-angle formula, we can express tan(2x) as:

tan(2x) = 2tan(x) / (1 - tan^2(x))

Let’s substitute u = tan(x), which means du/dx = sec^2(x). Therefore, du = sec^2(x) dx.

Step 2: Substitute and Integrate

We need to express tan(2x) in a form that’s easily integrable. Recall that:

tan(2x) = 2tan(x) / (1 - tan^2(x))

Substituting u = tan(x), we get:

tan(2x) = 2u / (1 - u^2)

However, to use this substitution effectively in the integral, let’s first find the integral of tan(2x) directly using a trigonometric substitution that fits the u = tan(x) substitution.

∫tan(2x) dx = ∫(2tan(x) / (1 - tan^2(x))) dx

Given u = tan(x), we transform the integral into:

∫(2u / (1 - u^2)) (1 / (1 + u^2)) du

Simplifying, we notice that:

∫(2u / (1 - u^2)) (1 / (1 + u^2)) du = ∫(2u / (1 - u^4)) du

Let’s focus on integrating 2u / (1 - u^4) with respect to u.

Step 3: Partial Fraction Decomposition

To integrate 2u / (1 - u^4), we can use partial fraction decomposition. Factor the denominator:

1 - u^4 = (1 - u^2)(1 + u^2)

So, we rewrite the integral as:

∫(2u / ((1 - u^2)(1 + u^2))) du

We can decompose 2u / ((1 - u^2)(1 + u^2)) into partial fractions as follows:

2u / ((1 - u^2)(1 + u^2)) = A / (1 - u^2) + B / (1 + u^2)

Multiplying both sides by (1 - u^2)(1 + u^2), we get:

2u = A(1 + u^2) + B(1 - u^2)

Choosing convenient values for u (e.g., u = 1, u = -1, u = 0), we can solve for A and B.

For u = 1, we get:

2 = 2A, so A = 1

For u = -1, we get:

-2 = 2B, so B = -1

Thus, the partial fraction decomposition is:

2u / ((1 - u^2)(1 + u^2)) = 1 / (1 - u^2) - 1 / (1 + u^2)

Step 4: Integrate the Partial Fractions

Now, integrate each partial fraction with respect to u:

∫(1 / (1 - u^2)) du - ∫(1 / (1 + u^2)) du

The first integral can be recognized as the integral for the hyperbolic tangent or solved through recognizing it as a form of the integral for the natural logarithm after a substitution. However, directly:

∫(1 / (1 - u^2)) du = (¹⁄₂)ln|(1 + u) / (1 - u)| + C1

The second integral is the arctan function:

∫(1 / (1 + u^2)) du = arctan(u) + C2

So, combining these results:

(¹⁄₂)ln|(1 + u) / (1 - u)| - arctan(u) + C

Where C = C1 - C2.

Step 5: Substitute Back to x

Recall that u = tan(x), so substituting back:

(¹⁄₂)ln|(1 + tan(x)) / (1 - tan(x))| - arctan(tan(x)) + C

Simplifying, we note that arctan(tan(x)) = x for x in the range of arctan, so:

(¹⁄₂)ln|(1 + tan(x)) / (1 - tan(x))| - x + C

This gives us the integral of tan(2x) in terms of x.

Conclusion

Finding the integral of tan(2x) involves expressing tan(2x) in terms of tan(x) and then using a substitution to simplify the integral into a form that can be solved using partial fractions. The solution involves natural logarithms and arctan functions, demonstrating the complex interplay between trigonometric functions and their integrals.

FAQ Section

In conclusion, integrating tan(2x) is a challenging problem that requires a deep understanding of trigonometric identities and integration techniques. By following the steps outlined above, one can derive the integral and gain insight into the complex relationships between trigonometric functions and their integrals.

Let’s move to advanced level problems similar to this for improving mathematical thinking skills and logical reasoning.