Question

The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx` is

विकल्प

• 0

• 2

• π

• 1

Answer 1

π

Explanation:

Let I = `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx`

`int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx + int_(- pi/2)^(pi/2)  dx = I_1 + int_(- pi/2)^(pi/2)  dx`

I₁ = `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx`

Let `f(x) = x^3 + x cos x + tan^5x`

`f(-x) = (-x)^3 + (-x) cos(-x) + tan^5 (-x)`

= `- x^3 - 2 cos x - tan^5x`

= `- (x^3 + 2 cos x + tan^5x)`

= `- f(x)`

I₁ = `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x)  dx`

= 0   .......`[because f(-x) = - f(x)]`

Since, if `f` is an odd function, 

Now, I = `I_1 + int_(- pi/2)^(pi/2)  dx`

= `0 + int_(- pi/2)^(pi/2)  dx`

= `[x]_(- pi/2)^(pi/2)`

= `pi/2 - (- pi/2)`

= π