Math Problem Statement

\int _0^{\frac{\pi }{4}}tan^4\theta :sec^4\theta :d\theta calc 2

Solution

We are tasked with evaluating the following integral:

I=0π4tan4(θ)sec4(θ)dθI = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \sec^4(\theta) \, d\theta

Step 1: Rewrite sec4(θ)\sec^4(\theta) in terms of tan(θ)\tan(\theta)

We can express sec2(θ)\sec^2(\theta) in terms of tan(θ)\tan(\theta) using the identity:

sec2(θ)=1+tan2(θ)\sec^2(\theta) = 1 + \tan^2(\theta)

Therefore, sec4(θ)\sec^4(\theta) becomes:

sec4(θ)=(sec2(θ))2=(1+tan2(θ))2\sec^4(\theta) = (\sec^2(\theta))^2 = (1 + \tan^2(\theta))^2

Now we substitute this into the integral:

I=0π4tan4(θ)(1+tan2(θ))2dθI = \int_0^{\frac{\pi}{4}} \tan^4(\theta) (1 + \tan^2(\theta))^2 \, d\theta

Step 2: Expand (1+tan2(θ))2(1 + \tan^2(\theta))^2

(1+tan2(θ))2=1+2tan2(θ)+tan4(θ)(1 + \tan^2(\theta))^2 = 1 + 2\tan^2(\theta) + \tan^4(\theta)

Now the integral becomes:

I=0π4tan4(θ)(1+2tan2(θ)+tan4(θ))dθI = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \left( 1 + 2\tan^2(\theta) + \tan^4(\theta) \right) \, d\theta

Step 3: Distribute tan4(θ)\tan^4(\theta)

I=0π4(tan4(θ)+2tan6(θ)+tan8(θ))dθI = \int_0^{\frac{\pi}{4}} \left( \tan^4(\theta) + 2\tan^6(\theta) + \tan^8(\theta) \right) \, d\theta

Step 4: Integrate term by term

Now we evaluate each term individually: I1=0π4tan4(θ)dθI_1 = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \, d\theta I2=0π42tan6(θ)dθI_2 = \int_0^{\frac{\pi}{4}} 2\tan^6(\theta) \, d\theta I3=0π4tan8(θ)dθI_3 = \int_0^{\frac{\pi}{4}} \tan^8(\theta) \, d\theta

To evaluate these integrals, we use reduction formulas for powers of tangent, or we can solve them using standard techniques such as substitution. I'll calculate these integrals now.

The individual integrals and the total result are as follows:

I1=0π4tan4(θ)dθ=π423I_1 = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \, d\theta = \frac{\pi}{4} - \frac{2}{3} I2=0π42tan6(θ)dθ=2615π2I_2 = \int_0^{\frac{\pi}{4}} 2\tan^6(\theta) \, d\theta = \frac{26}{15} - \frac{\pi}{2} I3=0π4tan8(θ)dθ=π476105I_3 = \int_0^{\frac{\pi}{4}} \tan^8(\theta) \, d\theta = \frac{\pi}{4} - \frac{76}{105}

Adding these together:

I=1235I = \frac{12}{35}

Thus, the value of the integral is 1235\frac{12}{35}.

Would you like further details on any step or a deeper explanation?

Here are 5 follow-up questions for more practice:

  1. How do you apply reduction formulas to integrals involving trigonometric powers?
  2. What is the general approach to solving integrals with products of powers of trigonometric functions?
  3. How would you compute 0π4sec6(θ)dθ\int_0^{\frac{\pi}{4}} \sec^6(\theta) \, d\theta?
  4. What substitution could simplify tann(θ)secm(θ)dθ\int \tan^n(\theta) \sec^m(\theta) \, d\theta?
  5. How does the identity sec2(θ)=1+tan2(θ)\sec^2(\theta) = 1 + \tan^2(\theta) simplify integration?

Tip: For higher powers of trigonometric functions, substitution with identities or recurrence relations can significantly simplify the process.

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Math Problem Analysis

Mathematical Concepts

Trigonometric Integrals
Power of Trigonometric Functions
Integration Techniques

Formulas

\( \sec^2(\theta) = 1 + \tan^2(\theta) \)
\( \sec^4(\theta) = (1 + \tan^2(\theta))^2 \)

Theorems

Trigonometric identities for powers of tangent and secant
Reduction formulas for trigonometric integrals

Suitable Grade Level

Grades 11-12 or Undergraduate Calculus