The value of the integral ∫₀^(π / 4) (x d x)/(sin ⁴(2 x)+cos ⁴(2 x)) equals :

Solution

Maths sir · 4 steps

Swap x for π/4-x: the denominator stays the same, so the x on top becomes a constant.

1. Use the King property. Replace x by π/4-x. Then sin 2x becomes cos 2x and cos 2x becomes sin 2x, so the denominator does not change. Add this new form of I to the original one, and the x in the numerator is replaced by π/4.

   2I = π/4∫₀^(π/4)dx/(sin⁴ 2x+cos⁴ 2x) ⇒ I = π/8·1/2∫₀^(π/2)dt/(sin⁴ t+cos⁴ t)

2. Change to tan. Here t = 2x, so dx = dt/2. Divide the top and bottom by cos⁴ t and put u = tan t. Since sec⁴ t dt = (1+u²) du, the integral becomes ∫₀^∞(1+u²)/(1+u⁴) du.
3. Standard substitution. Divide the top and bottom by u² to get (1+1/u²)/((u-1/u)²+2). Put z = u-1/u, which runs from -∞ to ∞. This gives ∫_(-∞)^∞dz/(z²+2) = π/√2.
4. So. I = π/8·1/2·π/√2 = π²/16√2 = (√2 π²)/32, which is option (c).

The idea to keep. When x multiplies a function that stays the same under x → a+b-x, add the two forms of the integral: x turns into (a+b)/2 and drops out of the integral.

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