The integral in the image is a double integral:

$\int_0^\pi \int_0^{2x+1} \sin(x) \, dy \, dx$

Steps to Solve:

1. Evaluate the inner integral: The inner integral is with respect to $y$, treating $\sin(x)$ as a constant since it does not depend on $y$: $\int_0^{2x+1} \sin(x) \, dy = \sin(x) \int_0^{2x+1} 1 \, dy = \sin(x) [y]_0^{2x+1}.$ Evaluate the limits: $\sin(x) [(2x+1) - 0] = \sin(x)(2x+1).$

2. Substitute back into the outer integral: Now, the integral becomes: $\int_0^\pi \sin(x)(2x+1) \, dx.$

3. Expand the integrand: $\int_0^\pi \sin(x)(2x+1) \, dx = \int_0^\pi 2x\sin(x) \, dx + \int_0^\pi \sin(x) \, dx.$

4. Solve each term:

   • First term: $\int_0^\pi 2x\sin(x) \, dx$ can be solved using integration by parts: Let $u = 2x$ and $dv = \sin(x) \, dx$. Then, $du = 2 \, dx$ and $v = -\cos(x)$. Applying integration by parts ($\int u \, dv = uv - \int v \, du$): $\int 2x\sin(x) \, dx = -2x\cos(x) \big|_0^\pi + \int_0^\pi 2\cos(x) \, dx.$ Evaluate each part:

     • For $-2x\cos(x) \big|_0^\pi$: At $x = \pi$, $\cos(\pi) = -1$, so $-2\pi(-1) = 2\pi$. At $x = 0$, $\cos(0) = 1$, so $-2(0)(1) = 0$. Hence, this term equals $2\pi$.
     • For $\int_0^\pi 2\cos(x) \, dx$: The integral of $2\cos(x)$ is $2\sin(x)$, so $2\sin(x) \big|_0^\pi = 2[\sin(\pi) - \sin(0)] = 0$. Thus, $\int_0^\pi 2x\sin(x) \, dx = 2\pi$.

   • Second term: $\int_0^\pi \sin(x) \, dx$: The integral of $\sin(x)$ is $-\cos(x)$, so: [ -\cos(x) \big|_0^\pi = -[\cos(\pi) - \cos(0)] = -[(-1) - 1] = 2. ]

5. Combine the results: $\int_0^\pi \sin(x)(2x+1) \, dx = 2\pi + 2 = 2(\pi + 1).$

Final Answer:

$\int_0^\pi \int_0^{2x+1} \sin(x) \, dy \, dx = 2(\pi + 1).$

Would you like further clarification or detailed steps? Let me know!

Related Questions:

1. What is the process of integration by parts?
2. How do you handle double integrals with variable limits?
3. What are the standard integrals of trigonometric functions like $\sin(x)$?
4. Why is $\sin(x)$ treated as a constant in the inner integral?
5. How do you verify the correctness of definite integrals?

Tip:

Always simplify the inner integral first before tackling the outer integral in a double integral problem.