Math Problem Statement

Evaluate the indefinite integrals below. No CAS. A. ∫132x cos(9 + 11x²) dx = B. ∫4 cos(x)(6 + sin(x))⁷ dx =

Solution

The image contains two indefinite integrals to solve:

A. 132xcos(9+11x2)dx\int 132x \cos(9 + 11x^2) \, dx

B. 4cos(x)(6+sin(x))7dx\int 4 \cos(x)(6 + \sin(x))^7 \, dx

Let's approach them one by one.

A. 132xcos(9+11x2)dx\int 132x \cos(9 + 11x^2) \, dx

For this, we can use u-substitution.

Let u=9+11x2u = 9 + 11x^2, so that du=22xdxdu = 22x \, dx.

Now, rewrite the integral:

132xcos(9+11x2)dx=13222cos(u)du=6cos(u)du\int 132x \cos(9 + 11x^2) \, dx = \int \frac{132}{22} \cos(u) \, du = 6 \int \cos(u) \, du

Now, integrate cos(u)\cos(u):

6sin(u)+C6 \sin(u) + C

Substitute back for uu:

6sin(9+11x2)+C6 \sin(9 + 11x^2) + C

B. 4cos(x)(6+sin(x))7dx\int 4 \cos(x)(6 + \sin(x))^7 \, dx

Here, we use u-substitution again.

Let u=6+sin(x)u = 6 + \sin(x), so du=cos(x)dxdu = \cos(x) \, dx.

Now, rewrite the integral:

4cos(x)(6+sin(x))7dx=4u7du\int 4 \cos(x)(6 + \sin(x))^7 \, dx = 4 \int u^7 \, du

Now, integrate u7u^7:

4u88+C=12u8+C4 \cdot \frac{u^8}{8} + C = \frac{1}{2} u^8 + C

Substitute back for uu:

12(6+sin(x))8+C\frac{1}{2} (6 + \sin(x))^8 + C


Final Answers:

  • A: 6sin(9+11x2)+C6 \sin(9 + 11x^2) + C
  • B: 12(6+sin(x))8+C\frac{1}{2} (6 + \sin(x))^8 + C

Would you like more details or help with anything else? Here are some related questions you might find useful:

  1. What other substitution techniques are useful in integration?
  2. Can we apply integration by parts to these integrals?
  3. How can we check the correctness of these indefinite integrals?
  4. How do we handle definite integrals with the same techniques?
  5. What are common mistakes in using substitution for integrals?

Tip: Remember to always carefully check the differential in u-substitution. If it's off by a factor, the whole integral will not work correctly!

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Math Problem Analysis

Mathematical Concepts

Integration
U-substitution
Trigonometric Integrals

Formulas

u-substitution formula: ∫f(g(x)) g'(x) dx = ∫f(u) du
Basic integral of cos(u): ∫cos(u) du = sin(u)
Basic integral of uⁿ: ∫uⁿ du = (uⁿ⁺¹)/(n+1)

Theorems

U-substitution
Basic integration techniques

Suitable Grade Level

Grades 11-12 (Calculus)