Final answer:
∫₀π / 6 ³θtanθ d θ = π² / 72
Step-by-step explanation:
The given integral is ∫₀π / 6 ³θtanθ d θ. To evaluate this integral, we use integration by parts, where ∫u dv = uv - ∫v du. Here, we assign u = θ³ and dv = tanθ dθ.
First, calculate du and v:
- Differentiating u with respect to θ, we get du = 3θ² dθ.
- Integrating dv, we get v = -ln|cosθ|.
Now, applying the integration by parts formula, uv - ∫v du becomes (θ³ * (-ln|cosθ|)) evaluated from 0 to π / 6 minus the integral of v du.
Evaluate the integral of v du by substituting u and v:
- ∫tanθ * θ² dθ = θ³ * (-ln|cosθ|) evaluated from 0 to π / 6 - ∫θ³ * (-ln|cosθ|) * 3θ² dθ.
Next, plug in the values:
- At the upper limit π / 6, the expression θ³ * (-ln|cosθ|) becomes (π / 6)³ * (-ln|cos(π / 6)|).
- Substituting θ = 0 results in 0, as sin(0) = 0 and ln|cos(0)| = ln(1) = 0.
- After computation, the final expression is ((π / 6)³ * (-ln|cos(π / 6)|)) - 0 - the integral of θ³ * (-ln|cosθ|) * 3θ² dθ.
Solving the integral ∫tanθ * θ² dθ yields the value of π² / 72 for the given integral ∫₀π / 6 ³θtanθ d θ.