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Question Number 58488 by Mr X pcx last updated on 23/Apr/19

l e t f (x) = \int \frac{d t}{x + c o s t + c o s (2 t)} (x r e a l)

1) f i n d a e x p l i c i t f o r m o f f (x)

2) d e t e r m i n e a l s o \int \frac{d t}{{(x + c o s t + c o s (2 t))}^{2}}

3) f i n d \int \frac{d t}{1 + c o s (t) + c o s (2 t)} a n d

\int \frac{d t}{{(3 + c o s (t) + c o s (2 t))}^{2}}

Commented by maxmathsup by imad last updated on 25/Apr/19

1) w e h a v e f (x) = \int \frac{d t}{x + c o s t + 2 {c o s}^{2} t - 1} l e t p u t c o s t = u a n d d e c o m p o s e

F (u) = \frac{1}{2 u^{2} + u + x - 1} \Rightarrow Δ = 1 - 8 (x - 1) = 9 - 8 x c a s e 1 9 - 8 x < 0 \Rightarrow n o r o o t s

a n d \frac{1}{2 u^{2} + u + x - 1} = \frac{1}{2 {u^{2} + \frac{u}{2} + \frac{x - 1}{2}}} = \frac{1}{2 {u^{2} + 2 \frac{1}{4} u + \frac{1}{16} + \frac{x - 1}{2} - \frac{1}{16}}}

= \frac{1}{2 {{(u + \frac{1}{2})}^{2} + \frac{8 x - 9}{16}}} \Rightarrow f (x) = \int \frac{d t}{2 {{(c o s t + \frac{1}{2})}^{2} + \frac{8 x - 9}{16}}}

c h a n g e m e n t c o s t + \frac{1}{2} = \frac{\sqrt{8 x - 9}}{4} u g i v e 4 c o s t + 2 = \sqrt{8 x - 9} u \Rightarrow

u = \frac{1}{\sqrt{8 x - 9}} (4 c o s t + 2) \Rightarrow d u = \frac{- 4 s i n t}{\sqrt{8 x - 9}} d t = \frac{- 4 \sqrt{1 - {c o s}^{2} t}}{\sqrt{8 x - 9}} d t \Rightarrow

\frac{d t}{d u} = - \frac{\sqrt{8 x - 9}}{4 \sqrt{1 - {(\frac{\sqrt{8 x - 9}}{4} u - \frac{1}{2})}^{2}}} \Rightarrow

f (x) = \int \frac{\sqrt{8 x - 9}}{8 \sqrt{1 - {(\frac{\sqrt{8 x - 9}}{4} u - \frac{1}{2})}^{2}} (\frac{8 x - 9}{16}) (1 + u^{2})} d u \dots b e c o n t i n u e d \dots

c a s e 2 i f 9 - 8 x > 0 \Rightarrow F (u) h a v e t w o p o l e s u_{1} = \frac{- 1 + \sqrt{9 - 8 x}}{4}

u_{2} = \frac{- 1 - \sqrt{9 - 8 x}}{4} a n d F (u) = \frac{1}{2 (u - u_{1}) (u - u_{2})}

= \frac{1}{2 (u_{1} - u_{2})} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} = \frac{1}{2 (\frac{\sqrt{9 - 8 x}}{2})} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} \Rightarrow

f (x) = \frac{1}{\sqrt{9 - 8 x}} {\int \frac{d t}{c o s t - u_{1}} - \int \frac{d t}{c o s t - u_{2}}} l e t f i n d \int \frac{d t}{c o s t - a}

c h . t a n (\frac{t}{2}) = u g i v e \int \frac{d t}{c o s t - a} = \int \frac{1}{\frac{1 - u^{2}}{1 + u^{2}} - a} \frac{2 d u}{1 + u^{2}}

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

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

Commented by maxmathsup by imad last updated on 26/Apr/19

\Rightarrow I = \int \frac{2 d u}{(1 + a) u^{2} + a - 1} = \frac{2}{1 + a} \int \frac{d u}{u^{2} + \frac{a - 1}{a + 1}} i f \frac{a - 1}{a + 1} > 0 w e d o t h e c h a n g e m e n t

u = \sqrt{\frac{a - 1}{a + 1}} t \Rightarrow I = \frac{2}{1 + a} \frac{a + 1}{a - 1} \int \frac{1}{t^{2} + 1} \sqrt{\frac{a - 1}{a + 1}} d t

= \frac{2}{\sqrt{a^{2} - 1}} a r c t a n (\sqrt{\frac{a + 1}{a - 1}} u) + c_{1} = \frac{2}{\sqrt{a^{2} - 1}} a r c t a n (\sqrt{\frac{a + 1}{a - 1}} t a n (\frac{t}{2})) + c_{1} \dots .