1-1-x-x-2-x-3-dx-2-1-tgx-sinx-dx-3-e-x-ln-1-1-x-2-dx-4-sinx-1-sinx-sin2x-dx-

Question Number 62227 by behi83417@gmail.com last updated on 17/Jun/19

1 . \int \sqrt{1 + x + x^{2} + x^{3}} dx = ?

2 . \int \frac{\sqrt{1 - tgx}}{sinx} dx = ?

3 . \int e^{x} . \ln (1 + \sqrt{1 + x^{2}}) dx = ?

4 . \int \frac{sinx}{1 + sinx + \sin 2 x} dx = ?

Commented by maxmathsup by imad last updated on 19/Jun/19

4) l e t I = \int \frac{s i n x}{1 + s i n x + s i n (2 x)} d x \Rightarrow I = \int \frac{s i n x}{1 + s i n x + 2 s i n x c o s x} d x

= \int \frac{d x}{\frac{1}{s i n x} + 1 + 2 c o s x} d x =_{t a n (\frac{x}{2}) = t} \int \frac{1}{\frac{1 + t^{2}}{2 t} + 1 + 2 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= \int \frac{4 t d t}{{(1 + t^{2})}^{2} + 6 t - 2 t^{3}} = \int \frac{4 t d t}{t^{4} + 2 t^{2} + 1 + 6 t - 2 t^{3}}

= \int \frac{4 t d t}{t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1} l e t d e c o m p o s e F (t) = \frac{4 t}{t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1}

t h e r o o t s o f P (x) = t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1 a r e t_{1} = - 1, t_{2} \sim - 0, 1795

t_{3} \sim 1, 5898 + 1, 7445 i (c o m p l e x)

t_{4} \sim 1, 5898 + 1, 7445 i (c o m p l e x) \Rightarrow

F (t) = \frac{4 t}{(t + 1) (t - t_{2}) (t - t_{3}) (t - {\bar{t}}_{3})} = \frac{4 t}{(t + 1) (t - t_{2}) (t^{2} - (t_{3} + {\bar{t}}_{3}) t + ∣ t_{3} ∣^{2})}

\Rightarrow F (t) = \frac{a}{t + 1} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} - α t + β} (α = t_{3} + {\bar{t}}_{3} a n d β =∣ t_{3} ∣^{2}) \Rightarrow

\int F (t) d t = a l n ∣ t + 1 ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} \int \frac{2 t - α + α}{t^{2} - α t + β} + \int \frac{d}{t^{2} - α t + β} d t

= a l n ∣ t + 1 ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} l n ∣ t^{2} - α t + β ∣ + (\frac{α c}{2} + d) \int \frac{d t}{t^{2} - α t + β}

\int \frac{d t}{t^{2} - α t + β} = \int \frac{d t}{t^{2} - 2 t \frac{α}{2} + β - \frac{α^{2}}{4}} = \int \frac{d t}{{(t - \frac{α}{2})}^{2} + \frac{4 β - α^{2}}{4}}

=_{t - \frac{α}{2} = \frac{\sqrt{4 β - α^{2}}}{2} u} \frac{1}{2} \int \frac{1}{\frac{4 β - α^{2}}{4} (1 + u^{2})} \sqrt{4 β - α^{2}} d u = \frac{2}{\sqrt{4 β - α^{2}}} a r c t a n (u) + C

= \frac{2}{\sqrt{4 β - α^{2}}} a r c t a n (\frac{2 t - α}{\sqrt{4 β^{2} - α^{2}}}) \Rightarrow

I = a l n ∣ 1 + t a n (\frac{x}{2}) ∣ + b l n ∣ t a n (\frac{x}{2}) - t_{2} ∣ + \frac{c}{2} l n ∣ {t a n}^{2} (\frac{x}{2}) - α t a n (\frac{x}{2}) + β ∣

+ (\frac{α c}{2} + d) \frac{2}{\sqrt{4 β - α^{2}}} a r c t a n (\frac{2 t a n (\frac{x}{2})}{\sqrt{4 β - α^{2}}}) + C r e s t t o c a l c u l a t e t h e v a l u e o f

a, b, c a n d d \dots .

Commented by behi83417@gmail.com last updated on 20/Jun/19

thanks in advance master proph . abdo!

nice and hard work .

Commented by prof Abdo imad last updated on 20/Jun/19

y o u a r e w e l c o m e s i r .