Q-Find-the-value-of-the-following-integral-I-0-pi-2-1-1-sin-4-x-cos-4-x-dx-

Question Number 186771 by mnjuly1970 last updated on 10/Feb/23

Q : Find the value of the following integral .

I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \sin^{4} (x) + \cos^{4} (x)} d x = ?

Answered by Ar Brandon last updated on 10/Feb/23

I = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + \sin^{4} x + \cos^{4} x} = \int_{0}^{\frac{π}{2}} \frac{\sec^{4} x}{\sec^{4} x + \tan^{4} x + 1} d x

= \int_{0}^{\frac{π}{2}} \frac{\tan^{2} x + 1}{{(\tan^{2} x + 1)}^{2} + \tan^{4} x + 1} d (\tan x)

= \int_{0}^{\infty} \frac{t^{2} + 1}{2 t^{4} + 2 t^{2} + 2} d t = \frac{1}{2} \int_{0}^{\infty} \frac{t^{2} + 1}{t^{4} + t^{2} + 1} d t

= \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{t^{2}}}{t^{2} + 1 + \frac{1}{t^{2}}} d t = \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 3} d t

= \frac{1}{2} \int_{- \infty}^{\infty} \frac{d u}{u^{2} + 3} = \frac{1}{2 \sqrt{3}} {[\arctan (\frac{u}{\sqrt{3}})]}_{- \infty}^{\infty}

= \frac{1}{2 \sqrt{3}} (\frac{π}{2} - - \frac{π}{2}) = \frac{1}{2 \sqrt{3}} (\frac{π}{2} + \frac{π}{2}) = \frac{π}{2 \sqrt{3}}

Commented by mnjuly1970 last updated on 11/Feb/23

t h a n k s s l o t s i r

Answered by integralmagic last updated on 10/Feb/23

= \frac{1}{2 \sqrt{3}} a r c t a n (\sqrt{3} / 2 t a n 2 x) ∣_{0}^{π / 2} = \frac{π}{2 \sqrt{3}}