find-0-pi-4-dt-1-cos-2-t-3-

Question Number 35058 by math khazana by abdo last updated on 14/May/18

f i n d \int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + {c o s}^{2} t)}^{3}}

Commented by abdo mathsup 649 cc last updated on 18/May/18

l e t p u t I = \int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + {c o s}^{2} t)}^{3}}

I = \int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + \frac{1 + c o s (2 t)}{2})}^{3}}

= 8 \int_{0}^{\frac{π}{4}} \frac{d t}{{(3 + c o s (2 t))}^{3}} =_{2 t = x} 8 \int_{0}^{\frac{π}{2}} \frac{1}{{(3 + c o s x)}^{3}} \frac{d x}{2}

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

g i v e I = 4 \int_{0}^{1} \frac{1}{{(3 + \frac{1 - t^{2}}{1 + t^{2}})}^{3}} \frac{2 d t}{1 + t^{2}}

I = 8 \int_{0}^{1} \frac{{(1 + t^{2})}^{2}}{{(3 + 3 t^{2} + 1 - t^{2})}^{3}} d t

I = 8 \int_{0}^{1} \frac{{(1 + t^{2})}^{2}}{{(4 + 2 t^{2})}^{3}} d t = \int_{0}^{1} \frac{{(1 + t^{2})}^{2}}{{(2 + t^{2})}^{3}} d t

I = \int_{0}^{1} \frac{t^{4} + 2 t^{2} + 1}{8 + 3 . 2^{2} t^{2} + 3.2 . t^{4} + t^{6}} d t

= \int_{0}^{1} \frac{t^{4} + 2 t^{2} + 1}{t^{6} + 6 t^{4} + 12 t^{2} + 8} d t l e t d e c o m p o s e

F (t) = \frac{t^{4} + 2 t^{2} + 1}{t^{6} + 6 t^{4} + 12 t^{2} + 8} \dots . . b e c o n t i n u e d \dots

Answered by MJS last updated on 16/May/18

\int \frac{d t}{{(1 + \cos^{2} (t))}^{3}} =

[\cos (t) = \frac{1}{\sec (t)}; \sec^{2} (t) = \tan^{2} (t) + 1]

= \int \frac{\sec^{2} (t) {(\tan^{2} (t) + 1)}^{2}}{{(\tan^{2} (t) + 2)}^{3}} d t =

[u = \tan (t) \to d t = \frac{d u}{\sec^{2} (t)}]

= \int \frac{{(u^{2} + 1)}^{2}}{{(u^{2} + 2)}^{3}} d u = \int \frac{d u}{u^{2} + 2} - 2 \int \frac{d u}{{(u^{2} + 2)}^{2}} + \int \frac{d u}{{(u^{2} + 2)}^{3}} =

\int \frac{d u}{u^{2} + 2} =

[v = \frac{\sqrt{2}}{2} u \to d u = d v \sqrt{2}]

= \frac{\sqrt{2}}{2} \int \frac{d v}{v^{2} + 1} = \frac{\sqrt{2}}{2} \arctan (v) = \frac{\sqrt{2}}{2} \arctan (\frac{\sqrt{2}}{2} u)

- 2 \int \frac{d u}{{(u^{2} + 2)}^{2}} =

[\int \frac{d u}{{({a u}^{2} + b)}^{n}} = \frac{u}{2 b (n - 1) {({a u}^{2} + b)}^{n - 1}} + \frac{2 n - 3}{2 b (n - 1)} \int \frac{d u}{{({a u}^{2} + b)}^{n - 1}}]

= - 2 (\frac{u}{4 (u^{2} + 2)} + \frac{1}{4} \int \frac{d u}{u^{2} + 2}) =

= - \frac{u}{2 (u^{2} + 2)} - \frac{\sqrt{2}}{4} \arctan (\frac{\sqrt{2}}{2} u)

\int \frac{d u}{{(u^{2} + 2)}^{3}} = \frac{u}{8 {(u^{2} + 2)}^{2}} + \frac{3}{8} \int \frac{d u}{{(u^{2} + 2)}^{2}} =

= \frac{u}{8 {(u^{2} + 2)}^{2}} + \frac{3}{8} (\frac{u}{4 (u^{2} + 2)} + \frac{\sqrt{2}}{8} \arctan (\frac{\sqrt{2}}{2} u)) =

= \frac{u}{8 {(u^{2} + 2)}^{2}} + \frac{3 u}{32 (u^{2} + 2)} + \frac{3 \sqrt{2}}{64} \arctan (\frac{\sqrt{2}}{2} u)

= \frac{19 \sqrt{2}}{64} \arctan (\frac{\sqrt{2}}{2} u) - \frac{13 u}{32 (u^{2} + 2)} + \frac{u}{8 {(u^{2} + 2)}^{2}} =

= \frac{19 \sqrt{2}}{64} \arctan (\frac{\sqrt{2}}{2} \tan (t)) - \frac{\tan (t) ({13 \tan}^{2} (t) + 22)}{32 {(\tan^{2} (t) + 2)}^{2}} + C

\underset{0}{\int^{\frac{π}{4}}} \frac{d t}{{(1 + \cos^{2} (t))}^{3}} = \frac{19 \sqrt{2}}{64} \arctan (\frac{\sqrt{2}}{2}) - \frac{35}{288}

Commented by abdo mathsup 649 cc last updated on 15/May/18

t h a n k y o u s i r M j s f o r t h i s h a r d w o r k \dots

Commented by MJS last updated on 16/May/18

{you}^{'} re welcome, it keeps my brains young

Commented by abdo mathsup 649 cc last updated on 16/May/18

a r e y o u a d o c t o r o r e n j e n e r s i r M j s \dots