Question

Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.

Answer 1

I = `int_0^π (x sin x)/(1 + cos^2x) dx`   ...(i)

= `int_0^π ((π - x) sin(π - x))/(1 + cos^2(π - x)) dx`   ...`["Using" int_0^a f(x) dx = int_0^a f(a - x) dx]`

= `int_0^π ((π - x) sin x)/(1 + cos^2x) dx`   ...(ii)

Adding (i) and (ii), we get,

2I = `int_0^π (x sin x)/(1 + cos^2x) dx + int_0^π ((π - x) sin x)/(1 + cos^2x) dx`

= `int_0^π ((x + π - x) sin x)/(1 + cos^2x) dx`

2I = `int_0^π (π sin x)/(1 + cos^2x) dx`

I = `π/2 int_0^π (sin x)/(1 + cos^2x) dx`

Let cos x = t, – sin x d x = dt

Also, x = 0, t = cos 0 = 1 and x = π, t = cos π = –1

I = `π/2 int_1^(-1) (-1)/(1 + t^2) dt`

I = `- π/2 [tan^-1 t]_1^(-1)`

= `- π/2 [tan^-1 (-1) - tan^(-1) (1)]`

= `- π/2 [- π/4 - π/4]`

= `π/2 xx (2π)/4`

I = `π^2/4`

Answer 2

I = `int_0^π (x sin x)/(1 + cos^2x) dx`   ...(i)

On Applying Property

`int_0^π f(x) dx = int_0^π f(π - x) dx`

I = `int_0^π ((π - x)sin(π - x))/(1 + cos^2(π - x)) dx`

or, I = `int_0^π ((π - x)sin x)/(1 + cos^2x) dx`   ...(ii)

Adding equation (i) and equation (ii), we get

2I = `int_0^π (π sin x)/(1 + cos^2x) dx`

I = `π/2 int_0^π (sin x)/(1 + cos^2x) dx`

Let cos x = t

– sin x dx = dt

sin x dx = – dt

When x = 0

t = cos 0 = 1

When x = π

t = cos π = –1

I = `- π/2 int_1^(-1) 1/(1 + t^2) dt`

Using Property

`int_a^b f(x)dx = -int_b^a f(x)dx`

I = `π/2 int_(-1)^1 1/(1 + t^2) dt`

I = `π/2 [tan^-1 t]_(-1)^1`

I = `π/2 [tan^-1 (1) - tan^-1 (-1)]`

= `π/2 [π/4 + (3π)/4]`

= `π/2 [+ (2π)/4]`

= `π^2/4`