Question

1) integral dx/sinx*cosx 2) integral dx/9x^2+16 3) x^2dx/корень 2x^3-5 4) 3xdx/5+x^2

Answer 1

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Answer 2

Let's solve each integral step by step:

1. ∫dx / (sinx * cosx)

To solve this integral, we can use the substitution method. Let's start by making a substitution:

Let u = sinx, then du = cosx dx

The integral becomes:

∫ du/u

Now, we can integrate with respect to u:

∫ du/u = ln|u| + C

Finally, substitute back u = sinx:

∫ dx / (sinx * cosx) = ln|sinx| + C

2. ∫ dx / (9x^2 + 16)

To solve this integral, we can use the method of partial fractions. The denominator cannot be factored over real numbers, so we'll use complex numbers:

9x^2 + 16 = (3x + 4i)(3x - 4i)

Now, we can express the original fraction as a sum of partial fractions:

∫ dx / (9x^2 + 16) = A/(3x + 4i) + B/(3x - 4i)

To find A and B, we can multiply through by the denominator and solve for the constants. After solving, we get:

A = (3i/32) B = (-3i/32)

Now the integral becomes:

∫ dx / (9x^2 + 16) = (3i/32) ∫ dx / (3x + 4i) - (3i/32) ∫ dx / (3x - 4i)

Integrating both parts:

= (3i/32) ln|3x + 4i| - (3i/32) ln|3x - 4i| + C

3. ∫ x^2 dx / √(2x^3 - 5)

To solve this integral, we can use substitution as well.

Let u = 2x^3 - 5, then du = 6x^2 dx

The integral becomes:

(1/6) ∫ du / √u

Now, we can integrate with respect to u:

(1/6) ∫ du / √u = (1/6) * 2√u + C = (√(2x^3 - 5))/3 + C

4. ∫ (3x dx) / (5 + x^2)

To solve this integral, we can also use substitution.

Let u = 5 + x^2, then du = 2x dx

The integral becomes:

(3/2) ∫ du / u

Now, we can integrate with respect to u:

(3/2) ∫ du / u = (3/2) ln|u| + C = (3/2) ln|5 + x^2| + C

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So, the solutions to the integrals are:

1. ∫ dx / (sinx * cosx) = ln|sinx| + C

2. ∫ dx / (9x^2 + 16) = (3i/32) ln|3x + 4i| - (3i/32) ln|3x - 4i| + C

3. ∫ x^2 dx / √(2x^3 - 5) = (√(2x^3 - 5))/3 + C

4. ∫ (3x dx) / (5 + x^2) = (3/2) ln|5 + x^2| + C