Question

B)  (e^8x+16/Sin^2 8x)dx

Answer 1

Ответ:

Пошаговое объяснение:

$\int {(e^x+\frac{16}{(sin(8x))^2} )} \, dx =\int {e^x} \, dx +16\int {(csc(8x))^2} \, dx =\frac{1}{8} e^x+(-\frac{16}{8} ctg(8x))+C=$

$=\frac{e^x}{8} -2ctg(8x) +C$

Answer 2

To find the integral of the function ∫ (e^(8x+16) / sin^2(8x)) dx, we can use trigonometric identities and substitution.

Let's start by simplifying the integrand:

∫ (e^(8x+16) / sin^2(8x)) dx

Now, we can use the identity for sin^2(8x):

sin^2(8x) = 1 - cos^2(8x)

So, our integral becomes:

∫ (e^(8x+16) / (1 - cos^2(8x))) dx

Next, we'll make a substitution to simplify this integral. Let's set:

u = cos(8x)

Now, calculate du:

du/dx = -8sin(8x)

dx = du / (-8sin(8x))

Now, substitute u and du into the integral:

∫ (e^(8x+16) / (1 - u^2)) * (-1/(8sin(8x))) du

Now, we need to express e^(8x+16) and sin(8x) in terms of u. Recall that:

e^(8x+16) = e^(8x) * e^16

sin(8x) = √(1 - cos^2(8x)) = √(1 - u^2)

So, our integral becomes:

∫ (-e^16 / 8) * (1 / (1 - u^2)) du

Now, we can simplify this further:

-1/8 * e^16 * ∫ (1 / (1 - u^2)) du

Now, we can use partial fraction decomposition to break this fraction into simpler fractions:

1 / (1 - u^2) = A / (1 - u) + B / (1 + u)

To find A and B, we can multiply both sides by the common denominator (1 - u)(1 + u):

1 = A(1 + u) + B(1 - u)

Now, we can solve for A and B. Let's first solve for A:

1 = A + Au + B - Bu

Now, group like terms:

(1 + B) = (A - B)u

Now, set the coefficients of u on both sides equal:

A - B = 0 (1)

Now, solve for B:

1 = A + B

Now, substitute the value of A from equation (1) into this equation:

1 = B + B

2B = 1

B = 1/2

Now, substitute B back into equation (1) to find A:

A - (1/2) = 0

A = 1/2

So, we have found that A = 1/2 and B = 1/2.

Now, we can rewrite our integral using these partial fractions:

-1/8 * e^16 * ∫ (1/2) * (1 / (1 - u)) + (1/2) * (1 / (1 + u)) du

Now, we can integrate each term separately:

-1/8 * e^16 * (1/2) * [ln|1 - u| - ln|1 + u|] + C

Now, substitute back for u:

-1/16 * e^16 * [ln|1 - cos(8x)| - ln|1 + cos(8x)|] + C

So, the integral of (e^(8x+16) / sin^2(8x)) dx is:

-1/16 * e^16 * [ln|1 - cos(8x)| - ln|1 + cos(8x)|] + C

Where C is the constant of integration.