Question

1) int dx/(sin^2x + cos^2x) 2) int x √(1 - 8x^2) × dx 3) int 3x × e^2x × dx 4) int (e^x × dx)/√(1 + e^x)

Answer 1

1) int dx/(sin^2x + cos^2x)=int dx=x+c
2) int x √(1 - 8x^2) × dx=int 1/2√(1-8x²)dx²=1/2*(-1/8)(1-8x²)^3/2+c=-1/16√(1-8x²)³+c
3) int 3x × e^2x × dx=
здесь нужно интегрировать по частям u=3x, du=3dx,  dv=e^2xdx, v=1/2e^2x и далее по формулt  int udv=uv-int vdu
  =3x*1/2e^2x-int 1/2e^2x*3dx=1,5xe^2x-0,75e^2x+c=0,75e^2x+c
4) int (e^x × dx)/√(1 + e^x)=int1/√(1+e^x)de^x=2√(1+e^x)+c

Answer 2

Sure, I can help you with these integrals. Here are the solutions for each of them:

1. ∫ dx / (sin^2x + cos^2x) This integral simplifies to: ∫ dx So, the integral of 1 with respect to x is simply: x + C, where C is the constant of integration.

2. ∫ x √(1 - 8x^2) dx This is an integral that can be solved using a trigonometric substitution. Let's make the substitution u = 2x, so du = 2dx. Then, we have: ∫ x √(1 - 8x^2) dx = (1/2) ∫ √(1 - u^2) du

Now, we can use a trigonometric substitution with u = sin(θ), so du = cos(θ)dθ and √(1 - u^2) = cos(θ): (1/2) ∫ cos^2(θ) dθ

You can integrate cos^2(θ) with respect to θ as follows: (1/2) ∫ (1 + cos(2θ))/2 dθ = (1/2) [θ/2 + (1/4)sin(2θ)] + C

Now, we need to go back to the x variable: θ = sin^(-1)(u) = sin^(-1)(2x) sin(2θ) = 2sin(θ)cos(θ) = 2(2x)√(1 - 4x^2) = 4x√(1 - 4x^2)

So, the final result is: (1/2) [sin^(-1)(2x)/2 + (1/4)(4x√(1 - 4x^2))] + C (1/4)sin^(-1)(2x) + x√(1 - 4x^2) + C

3. ∫ 3x e^(2x) dx This integral can be solved using integration by parts. Let u = 3x and dv = e^(2x) dx. Then, du = 3 dx and v = (1/2)e^(2x). The integration by parts formula is ∫ u dv = uv - ∫ v du, so: ∫ 3x e^(2x) dx = (3x)(1/2)e^(2x) - ∫ (1/2)e^(2x) * 3 dx

Simplify this: (3/2)x e^(2x) - (3/2)∫ e^(2x) dx

Now, integrate the remaining integral: (3/2)x e^(2x) - (3/2) * (1/2)e^(2x) + C

Combine the terms: (3/2)x e^(2x) - (3/4)e^(2x) + C

4. ∫ (e^x dx) / √(1 + e^x) This integral can be solved using a trigonometric substitution. Let's make the substitution u = √(1 + e^x), so du = (1/2)e^x√(1 + e^x) dx: ∫ (e^x dx) / √(1 + e^x) = 2∫ du

Now, you can easily integrate with respect to u: 2u + C

Finally, substitute back for u: 2√(1 + e^x) + C