Question Statement

Evaluate the integral:

$\int y\sin 2ydy$

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Background and Explanation

This problem involves integrating the product of an algebraic function ($y$) and a trigonometric function ($\sin 2y$). When faced with integrals of products, integration by parts is the standard technique, particularly when one function is easily differentiable (like $y$) and the other is easily integrable (like $\sin 2y$).

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Solution

We begin by denoting the integral as $I$:

$I=\int y\sin 2ydy$

Step 1: Choose $u$ and $dv$

Using the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential), we select:

• $u=y$ (algebraic function, easy to differentiate)
• $dv=\sin 2ydy$ (trigonometric function, easy to integrate)

Step 2: Compute $du$ and $v$

Differentiating $u$: $du=\frac{d}{dy}(y)dy=1\cdot dy=dy$

Integrating $dv$: $v=\int \sin 2ydy=\frac{-\cos 2y}{2}$

Step 3: Apply the integration by parts formula

The formula states: $\int udv=uv-\int vdu$

Substituting our values: $I=y\cdot \left(\frac{-\cos 2y}{2}\right)-\int \left(\frac{-\cos 2y}{2}\right)\cdot dy$

Step 4: Simplify the expression

$I=\frac{-y\cos 2y}{2}+\frac{1}{2}\int \cos 2ydy$

The negative signs cancel in the second term, and we factor out the constant $\frac{1}{2}$.

Step 5: Evaluate the remaining integral

$\int \cos 2ydy=\frac{\sin 2y}{2}$

Step 6: Combine all terms and add the constant of integration

$I=\frac{-y\cos 2y}{2}+\frac{1}{2}\cdot \frac{\sin 2y}{2}+c$

$I=\frac{-y}{2}\cos 2y+\frac{1}{4}\sin 2y+c$

Therefore, the final answer is:

$\boxed{\int y\sin 2ydy=-\frac{y}{2}\cos 2y+\frac{1}{4}\sin 2y+c}$

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Key Formulas or Methods Used

• Integration by parts formula: $\int udv=uv-\int vdu$
• Standard trigonometric integrals:
  • $\int \sin (ax)dx=-\frac{\cos (ax)}{a}+C$
  • $\int \cos (ax)dx=\frac{\sin (ax)}{a}+C$
• LIATE rule: Priority order for choosing $u$ (Logarithmic $\to$ Inverse $\to$ Algebraic $\to$ Trigonometric $\to$ Exponential)

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Summary of Steps

1. Identify the method: Recognize this as a product requiring integration by parts.
2. Select $u$ and $dv$: Set $u=y$ and $dv=\sin 2ydy$ following the LIATE priority.
3. Differentiate and integrate: Calculate $du=dy$ and $v=-\frac{\cos 2y}{2}$.
4. Apply the formula: Substitute into $\int udv=uv-\int vdu$.
5. Simplify: Handle the negative signs and factor out constants to get $\frac{-y\cos 2y}{2}+\frac{1}{2}\int \cos 2ydy$.
6. Integrate the remainder: Evaluate $\int \cos 2ydy=\frac{\sin 2y}{2}$.
7. Combine terms: Multiply and add the constant of integration to obtain the final result.