1-find-f-x-0-pi-4-sint-2-x-cos-2t-dt-2-find-g-x-0-pi-4-sint-sin-2t-2-x-cos-2t-2-dx-3-find-the-value-of-0-pi-4-sint-2-3-cos-2t-dt-and-0-pi-4-sin-t-

Question Number 49635 by maxmathsup by imad last updated on 08/Dec/18

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

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

3) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 + 3 c o s (2 t)} d t a n d \int_{0}^{\frac{π}{4}} \frac{s i n (t) s i n (2 t)}{{(2 + 3 c o s (2 t))}^{2}} d t

Commented by Abdo msup. last updated on 15/Dec/18

1) w e h a v e f (x) = \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 + x (2 {c o s}^{2} t - 1)} d t

= \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 {x c o s}^{2} t + 2 - x} d t =_{c o s t = u} \int_{1}^{\frac{1}{\sqrt{2}}} \frac{- d u}{2 {x u}^{2} + 2 - x}

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

i f \frac{2 - x}{2 x} > 0 w e d o t h e c h a n g e m e n t u = \sqrt{\frac{2 - x}{2 x}} α

f (x) = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2} \sqrt{\frac{2 x}{2 - x}}}^{\sqrt{\frac{2 x}{2 - x}}} \frac{1}{\frac{2 - x}{2 x} (1 + α^{2})} \sqrt{\frac{2 - x}{2 x}} d α

= \frac{1}{2 - x} \sqrt{\frac{2 - x}{2 x}} {[a r c t a n (α)]}_{\frac{\sqrt{2}}{2} \sqrt{\frac{2 x}{2 - x}}}^{\sqrt{\frac{2 x}{2 - x}}}

f (x) = \frac{1}{\sqrt{(2 - x) 2 x}} {a r c t a n (\sqrt{\frac{2 x}{2 - x}}) - a r c t a n (\frac{\sqrt{2}}{2} \sqrt{\frac{2 x}{2 - x}})}

i f \frac{2 - x}{2 x} < 0 w e g e t f (x) = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2}}^{1} \frac{d u}{u^{2} - \frac{x - 2}{2 x}} a n d w e d o

t h e c h a n g e m e n t u = \sqrt{\frac{x - 2}{2 x}} α a n d t h a t l e a d t o

f (x) = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}}}^{\sqrt{\frac{x - 2}{2 x}}} \frac{1}{\frac{x - 2}{2 x} (α^{2} - 1)} \sqrt{\frac{x - 2}{2 x}} d α

= \frac{1}{\sqrt{(x - 2) 2 x}} \frac{1}{2} \int_{. .}^{. .} {\frac{1}{α - 1} - \frac{1}{α + 1}} d α

= \frac{1}{2 \sqrt{2 x (x - 2)}} {[l n ∣ \frac{α - 1}{α + 1} ∣]}_{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}}}^{\sqrt{\frac{x - 2}{2 x}}}

= \frac{1}{2 \sqrt{2 x (x - 2)}} {l n ∣ \frac{\sqrt{\frac{x - 2}{2 x}} - 1}{\sqrt{\frac{x - 2}{2 x}} + 1} ∣ - l n ∣ \frac{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}} - 1}{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}} + 1} ∣ .

Answered by Smail last updated on 09/Dec/18

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

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

u = c o s t \Rightarrow d u = - s i n t d t

f (x) = - \int_{1}^{\sqrt{2} / 2} \frac{d u}{2 + 2 {x u}^{2} - x}

= - \int_{1}^{1 / \sqrt{2}} \frac{d u}{(2 - x) (\frac{2 {x u}^{2}}{2 - x} + 1)}

y = \sqrt{\frac{2 x}{2 - x}} u \Rightarrow d u = \sqrt{\frac{2 - x}{2 x}} d y

f (x) = - \frac{1}{2 - x} \times \sqrt{\frac{2 - x}{2 x}} \int_{\sqrt{2 x / (2 - x)}}^{\sqrt{x / (2 - x)}} \frac{d y}{y^{2} + 1}

= - \frac{1}{\sqrt{2 x (2 - x)}} [{t a n}^{- 1} (\sqrt{\frac{x}{2 - x}}) - {t a n}^{- 1} (\sqrt{\frac{2 x}{2 - x}})]

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