calculate-0-pi-4-dx-cos-3-x-sin-3-x-

Question Number 34219 by abdo imad last updated on 03/May/18

c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d x}{{c o s}^{3} x + {s i n}^{3} x}

Commented by abdo imad last updated on 31/May/18

I = \int_{0}^{\frac{π}{4}} \frac{d x}{(c o s x + s i n x) ({c o s}^{2} x - c o s x s i n x + {s i n}^{2} x)}

= \int_{0}^{\frac{π}{4}} \frac{d x}{(c o s x + s i n x) (1 - c o s x s i n x)}

= \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{2} c o s (\frac{π}{4} - x) (1 - \frac{1}{2} s i n (2 x))}

=_{\frac{π}{4} - x = t} \int_{\frac{π}{4}}^{- \frac{π}{4}} \frac{- d t}{\sqrt{2} c o s t (1 - \frac{1}{2} s i n (2 (\frac{π}{4} - t))}

= \sqrt{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{d t}{c o s (t) (2 - s i n (\frac{π}{2} - 2 t))}

= \sqrt{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{d t}{c o s (t) (2 - c o s (2 t))}

= 2 \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{d t}{c o s t (2 - 2 {c o s}^{2} t + 1)}

= 2 \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{d t}{c o s t (3 - 2 {c o s}^{2} t)} c h a n g e m e n t t a n (\frac{t}{2}) = x

g i v e

I = 2 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{1}{\frac{1 - x^{2}}{1 + x^{2}} {3 - 2 {(\frac{1 - x^{2}}{1 + x^{2}})}^{2}}} \frac{2 d x}{1 + x^{2}}

= 4 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{d x}{(1 - x^{2}) {\frac{3 {(1 + x^{2})}^{2} - 2 {(1 - x^{2})}^{2}}{{(1 + x^{2})}^{2}}}}

= 4 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{{(1 + x^{2})}^{2}}{(1 - x^{2}) {3 {(1 + x^{2})}^{2} - 2 {(1 - x^{2})}^{2}}} d x

= 4 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{x^{4} + 2 x^{2} + 1}{(1 - x^{2}) (3 x^{4} + 6 x^{2} + 3 - 2 x^{4} + 4 x^{2} - 2}} d x

= 4 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{x^{4} + 2 x^{2} + 1}{(1 - x^{2}) (x^{4} + 10 x^{2} + 1)} d x \dots . b e c o n t i n u e d \dots