Question

∫0π (xtanx) dx/(secx + cosx) = ?

Answer 1

1. Simplify the Integrand:
∫₀^π (x tan(x)) / (sec(x) + cos(x)) dx

We know that tan(x) = sin(x) / cos(x) and sec(x) = 1 / cos(x). Substituting these into the integral:

∫₀^π x * (sin(x) / cos(x)) / ((1 / cos(x)) + cos(x)) dx

To simplify the denominator, find a common denominator:

(1 / cos(x)) + cos(x) = (1 + cos²(x)) / cos(x)

Substitute this back into the integral:

∫₀^π x * (sin(x) / cos(x)) / ((1 + cos²(x)) / cos(x)) dx

Simplify by multiplying by the reciprocal of the denominator:

∫₀^π x * (sin(x) / cos(x)) * (cos(x) / (1 + cos²(x))) dx

The cos(x) terms cancel:

∫₀^π x * sin(x) / (1 + cos²(x)) dx

2. Apply the Property of Definite Integrals:
We can use the property ∫₀^a f(x) dx = ∫₀^a f(a - x) dx.

Let I = ∫₀^π x * sin(x) / (1 + cos²(x)) dx

Then, I = ∫₀^π (π - x) * sin(π - x) / (1 + cos²(π - x)) dx

Since sin(π - x) = sin(x) and cos(π - x) = -cos(x), we have:

I = ∫₀^π (π - x) * sin(x) / (1 + (-cos(x))²) dx

I = ∫₀^π (π - x) * sin(x) / (1 + cos²(x)) dx

3. Add the Two Integrals:
Add the original integral and the transformed integral:

2I = ∫₀^π x * sin(x) / (1 + cos²(x)) dx + ∫₀^π (π - x) * sin(x) / (1 + cos²(x)) dx

2I = ∫₀^π (x + π - x) * sin(x) / (1 + cos²(x)) dx

2I = ∫₀^π π * sin(x) / (1 + cos²(x)) dx

2I = π ∫₀^π sin(x) / (1 + cos²(x)) dx

4. Solve the Integral:
Let u = cos(x). Then du = -sin(x) dx.

When x = 0, u = cos(0) = 1.
When x = π, u = cos(π) = -1.

2I = π ∫₁^(-1) -du / (1 + u²)

2I = π ∫(-1)^1 du / (1 + u²)

2I = π [arctan(u)]_(-1)^1

2I = π [arctan(1) - arctan(-1)]

2I = π [π/4 - (-π/4)]

2I = π [π/2]

2I = π²/2

5. Find I:
I = π²/4

Therefore, the value of the integral is π²/4.