Understanding and Solving the Integral of xsinx/1-cosx

The integral of xsinx/1-cosx is an interesting problem in calculus. Such integrals often involve complex trigonometric identities and require a deep understanding of integration techniques.

Integral of xsinx/1-cosx

First, let's recall and simplify the given integral:

I ∫(xsinx)/(1 - cosx) dx

Using the trigonometric identity sinx(1 - cosx) 2sin(x/2)cos(x/2), we can rewrite the integral as:

I ∫x(2sin(x/2)cos(x/2))/(2sin^2(x/2)) dx ∫(x/2 sec^2(x/2) - cot(x/2)) dx

From here, we have:

I -∫(cot(x/2) - x/2 sec^2(x/2)) dx

Upon integrating, we get:

I -x cot(x/2) C

Maximum and Minimum Values of sinx

The maximum value of sinx is one, which occurs at the following points:

x π/2 2kπ, where k is an integer.

At these points, cosx is zero. When sinx equals zero, the value of cosx is either 1 or -1, depending on the value of x. This occurs at:

x π 2kπ, where k is an integer.

Solving the Equation sinx-cosx1

Let's solve the equation sinx - cosx 1 using standard trigonometric identities:

sinx - cosx 1

Squaring both sides, we obtain:

sin^2x cos^2x - 2sinx cosx 1

Since sin^2x cos^2x 1, we get:

1 - 2sinx cosx 1

Thus,

2sinx cosx 0

This implies:

sin2x 0

Therefore,

2x sin^-1(0)

Hence,

x π/2 kπ, where k is an integer.

General Solutions for sinx-cosx1

To find the general solutions for the equation:

sinx - cosx 1

We substitute sinx y and remember that:

cosx sqrt{1 - sin^2x}

Substituting and simplifying, we get:

y 1/sqrt{1 - y^2}

Rearranging, we obtain:

sqrt{1 - y^2} y - 1

Squaring both sides:

1 - y^2 (y - 1)^2

This simplifies to:

y^2 - 2y 1 1 - y^2

2y^2 - 2y 0

2y(y - 1) 0

Hence, the solutions are:

y_1 0 and y_2 1

Reversing the substitution, we find the three possible solutions in the interval [0, 2π]:

x 0, x π/2, and x π

However, only x π and x π/2 satisfy the original equation.

Finally, considering the periodicity of trigonometric functions, we have:

x 2kπ and x 2k(π/2)

where k ∈ mathbb{N}.