Math Problem Statement

(1 point) Evaluate the integral:

∫π/3 0 ⟨5sin^4(t)cos(t),−3cos^2(t)sin(t),3tan^2(t)sec^2(t)⟩ dt

Solution

To evaluate the vector integral

0π35sin4(t)cos(t),3cos2(t)sin(t),3tan2(t)sec2(t)dt,\int_{0}^{\frac{\pi}{3}} \langle 5\sin^4(t)\cos(t), -3\cos^2(t)\sin(t), 3\tan^2(t)\sec^2(t) \rangle \, dt,

we can break it into three separate scalar integrals for each component of the vector.

First component: 0π35sin4(t)cos(t)dt\int_0^{\frac{\pi}{3}} 5\sin^4(t)\cos(t) \, dt

Using the substitution u=sin(t)u = \sin(t), which gives du=cos(t)dtdu = \cos(t) \, dt, we can rewrite the integral as:

0π35sin4(t)cos(t)dt=0sin(π3)5u4du=0325u4du.\int_0^{\frac{\pi}{3}} 5\sin^4(t)\cos(t) \, dt = \int_0^{\sin(\frac{\pi}{3})} 5u^4 \, du = \int_0^{\frac{\sqrt{3}}{2}} 5u^4 \, du.

This evaluates to:

5u55032=(32)5=3332.\frac{5u^5}{5} \Big|_0^{\frac{\sqrt{3}}{2}} = \left(\frac{\sqrt{3}}{2}\right)^5 = \frac{3\sqrt{3}}{32}.

Second component: 0π33cos2(t)sin(t)dt\int_0^{\frac{\pi}{3}} -3\cos^2(t)\sin(t) \, dt

Using the substitution u=cos(t)u = \cos(t), with du=sin(t)dtdu = -\sin(t) \, dt, we get:

0π33cos2(t)sin(t)dt=1123u2du=[u3]112=181=78.\int_0^{\frac{\pi}{3}} -3\cos^2(t)\sin(t) \, dt = \int_1^{\frac{1}{2}} 3u^2 \, du = \left[ u^3 \right]_1^{\frac{1}{2}} = \frac{1}{8} - 1 = -\frac{7}{8}.

Third component: 0π33tan2(t)sec2(t)dt\int_0^{\frac{\pi}{3}} 3\tan^2(t)\sec^2(t) \, dt

Recognize that tan2(t)sec2(t)dt\int \tan^2(t)\sec^2(t) \, dt is just a standard integral:

tan2(t)sec2(t)dt=tan3(t)3.\int \tan^2(t)\sec^2(t) \, dt = \frac{\tan^3(t)}{3}.

Evaluating this from 00 to π3\frac{\pi}{3}:

[tan3(t)]0π3=tan3(π3)=(3)3=33.\left[ \tan^3(t) \right]_0^{\frac{\pi}{3}} = \tan^3\left(\frac{\pi}{3}\right) = \left(\sqrt{3}\right)^3 = 3\sqrt{3}.

Final Result

Thus, the evaluated integral is:

0π35sin4(t)cos(t),3cos2(t)sin(t),3tan2(t)sec2(t)dt=3332,78,33.\int_{0}^{\frac{\pi}{3}} \langle 5\sin^4(t)\cos(t), -3\cos^2(t)\sin(t), 3\tan^2(t)\sec^2(t) \rangle \, dt = \langle \frac{3\sqrt{3}}{32}, -\frac{7}{8}, 3\sqrt{3} \rangle.

Would you like further explanation or details?

Here are some related questions to expand your understanding:

  1. How do you solve integrals involving trigonometric identities?
  2. What is the role of substitution in evaluating trigonometric integrals?
  3. Can you explain how trigonometric functions behave over different intervals?
  4. How can we simplify powers of trigonometric functions before integration?
  5. What are common techniques for solving vector integrals?

Tip: Always check if substitution simplifies the integral before using trigonometric identities!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometric Integrals
Vector Calculus
Substitution Method
Trigonometric Identities

Formulas

∫ sin^4(t)cos(t) dt
∫ cos^2(t)sin(t) dt
∫ tan^2(t)sec^2(t) dt

Theorems

Substitution Rule
Basic Integral Properties of Trigonometric Functions

Suitable Grade Level

Undergraduate Level (Calculus II or III)