The value of ∫_(- π/2)^(π/2) ( (1 + sin ²x)/(1 + π ˢⁱⁿˣ) ) dx is

Solution

Maths sir · 4 steps

Replace x by −x and add the two integrals: the awkward πˢⁱⁿˣ denominator disappears.

1. Use the property. On a symmetric interval, ∫₋ₐᵃ f(x) dx = ∫₋ₐᵃ f(-x) dx. Call the integral I. Since sin(-x) = -sin x, the denominator becomes 1 + π⁻ˢⁱⁿˣ.
2. Rewrite the denominator. Multiply top and bottom by πˢⁱⁿˣ. That gives I = ∫_(-π/2)^(π/2) πˢⁱⁿˣ(1+sin² x)/(1+πˢⁱⁿˣ) dx.

   2I = ∫_(-π/2)^(π/2) (1+πˢⁱⁿˣ)(1+sin² x)/(1+πˢⁱⁿˣ) dx = ∫_(-π/2)^(π/2) (1+sin² x) dx

3. Evaluate. ∫_(-π/2)^(π/2) 1 dx = π. Since sin² x = (1-cos 2x)/2, we get ∫_(-π/2)^(π/2) sin² x dx = π/2. So 2I = π + π/2 = 3π/2.
4. So. I = 3π/4, which is option (c).

The idea to keep. When the denominator is 1 + aᵍ⁽ˣ⁾ with g odd and the limits are symmetric, change x to -x and add. The denominator cancels and you are left with the plain numerator.

Answer from the NTA final answer key. Solution by PrepWiser.