Math Problem Statement

Determine ∫y dx in each of the following cases: y = 2 sec²(π/3 x) tan(π/3 x), y = sin(-6x) sin(3x), y = sin²(3/4 x), y = (4x³ - 6x² - 10x + 9)/(2x - 1), y = cos(-x)e^(-4x). Also solve a partial fraction problem y = (x - 4)/(2x² - 5x - 3).

Solution

The image contains a series of integration problems (4.1 to 4.5). Let me solve them step by step:


4.1.1

y=2sec2(π3x)tan(π3x)y = 2 \sec^2\left(\frac{\pi}{3}x\right) \tan\left(\frac{\pi}{3}x\right) To integrate ydx\int y \, dx:

  • Use the identity: ddx[sec2(u)]=tan(u)sec2(u)dudx\frac{d}{dx}[\sec^2(u)] = \tan(u) \cdot \sec^2(u) \cdot \frac{du}{dx}.
  • Let u=π3xu = \frac{\pi}{3}x. Then dudx=π3\frac{du}{dx} = \frac{\pi}{3}.

Substitute: 2sec2(π3x)tan(π3x)dx=21π3sec2(u)tan(u)du=6πsec2(u)+C.\int 2 \sec^2\left(\frac{\pi}{3}x\right) \tan\left(\frac{\pi}{3}x\right) \, dx = 2 \cdot \frac{1}{\frac{\pi}{3}} \int \sec^2(u) \tan(u) \, du = \frac{6}{\pi} \sec^2(u) + C.

Final result: 6πsec2(π3x)+C.\frac{6}{\pi} \sec^2\left(\frac{\pi}{3}x\right) + C.


4.1.2

y=sin(6x)sin(3x)y = \sin(-6x)\sin(3x)

Use the product-to-sum formula: sin(A)sin(B)=12[cos(AB)cos(A+B)].\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)].

So: [ \sin(-6x)\sin(3x) = \frac{1}{2}[\cos(-6x - 3x) - \cos(-6x + 3x)] = \frac{1}{2}[\cos(-9x) - \cos(-3x)]. ]

Now integrate term-by-term: ydx=12cos(9x)dx12cos(3x)dx.\int y \, dx = \frac{1}{2} \int \cos(-9x) \, dx - \frac{1}{2} \int \cos(-3x) \, dx.

cos(kx)dx=sin(kx)k.\int \cos(kx) \, dx = \frac{\sin(kx)}{k}.

Substitute: cos(9x)dx=sin(9x)9,cos(3x)dx=sin(3x)3.\int \cos(-9x) \, dx = -\frac{\sin(9x)}{9}, \quad \int \cos(-3x) \, dx = -\frac{\sin(3x)}{3}.

Final result: sin(9x)18+sin(3x)6+C.-\frac{\sin(9x)}{18} + \frac{\sin(3x)}{6} + C.


4.1.3

y=sin2(34x)y = \sin^2\left(\frac{3}{4}x\right)

Use the identity: sin2(u)=12[1cos(2u)].\sin^2(u) = \frac{1}{2}[1 - \cos(2u)].

Let u=34xu = \frac{3}{4}x. Then: sin2(34x)=12[1cos(32x)].\sin^2\left(\frac{3}{4}x\right) = \frac{1}{2}\left[1 - \cos\left(\frac{3}{2}x\right)\right].

Integrate term-by-term: ydx=121dx12cos(32x)dx.\int y \, dx = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos\left(\frac{3}{2}x\right) \, dx.

1dx=x,cos(32x)dx=sin(32x)32=23sin(32x).\int 1 \, dx = x, \quad \int \cos\left(\frac{3}{2}x\right) \, dx = \frac{\sin\left(\frac{3}{2}x\right)}{\frac{3}{2}} = \frac{2}{3} \sin\left(\frac{3}{2}x\right).

Final result: 12x13sin(32x)+C.\frac{1}{2}x - \frac{1}{3}\sin\left(\frac{3}{2}x\right) + C.


4.1.4

y=4x36x210x+92x1y = \frac{4x^3 - 6x^2 - 10x + 9}{2x - 1}

This requires polynomial long division:

  1. Perform division: Divide 4x36x210x+94x^3 - 6x^2 - 10x + 9 by 2x12x - 1.
  2. Integrate the result.

4.1.5

y=cos(x)e4xy = \cos(-x)e^{-4x}

Here, the negative sign in cos(x)\cos(-x) doesn't change the result (cos(x)=cos(x)\cos(-x) = \cos(x)): y=cos(x)e4x.y = \cos(x)e^{-4x}.

Use integration by parts: Let: u=cos(x),dv=e4xdx.u = \cos(x), \quad dv = e^{-4x}dx.

Then: du=sin(x)dx,v=14e4x.du = -\sin(x)dx, \quad v = -\frac{1}{4}e^{-4x}.

Integration by parts formula: udv=uvvdu.\int u dv = uv - \int v du.

Substitute: cos(x)e4xdx=14cos(x)e4x14e4x(sin(x))dx.\int \cos(x)e^{-4x}dx = -\frac{1}{4}\cos(x)e^{-4x} - \int -\frac{1}{4}e^{-4x}(-\sin(x))dx.

Solve recursively.


Let me know if you want solutions for specific cases, or for me to detail partial fraction decomposition for 4.1.44.1.4. 😊

Follow-up Questions

  1. Do you want further breakdown of integration techniques like substitution or parts?
  2. Should I explain polynomial long division for 4.1.44.1.4?
  3. Would you like to see the detailed steps for 4.1.54.1.5 (recursive integration)?
  4. Do you need help with simplifying trigonometric identities?
  5. Are there any specific steps where you'd like clarification?

Tip:

Always check if trigonometric integrals can be simplified using identities before proceeding to solve. It often makes the process faster!

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Math Problem Analysis

Mathematical Concepts

Trigonometric identities
Integration by substitution
Integration by parts
Partial fractions decomposition
Polynomial long division

Formulas

∫sec²(u)tan(u) du = sec²(u)
sin(A)sin(B) = 1/2[cos(A-B) - cos(A+B)]
sin²(u) = 1/2[1 - cos(2u)]
∫cos(kx) dx = sin(kx)/k
∫u dv = uv - ∫v du

Theorems

Trigonometric product-to-sum identities
Integration by parts theorem
Fundamental theorem of calculus

Suitable Grade Level

College-level Calculus