Integrate-the-following-1-a-x-x-dx-2-dx-sin-x-3-2-cos-x-

Question Number 84234 by niroj last updated on 10/Mar/20

Integrate the following :

1 . \int \sqrt{\frac{a + x}{x}} dx

2 . \int \frac{dx}{\sin x (3 + 2 \cos x)}

Commented by mathmax by abdo last updated on 10/Mar/20

I = \int \sqrt{\frac{a + x}{x}} d x \Rightarrow I = \int \frac{\sqrt{a + x}}{\sqrt{x}} d x =_{\sqrt{x} = t} \int \frac{\sqrt{a + t^{2}}}{t} (2 t) d t

= 2 \int \sqrt{a + t^{2}} d t =_{t = \sqrt{a} s h (u)} 2 \int \sqrt{a} c h (u) \sqrt{a} c h (u) d u

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

= a \int (1 + c h (2 u)) d u = a u + \frac{a}{2} s h (2 u) + C

= a u + a s h (u) c h (u) + C b u t u = a r g s h (\frac{t}{\sqrt{a}}) = a r g s h (\sqrt{\frac{x}{a}})

= l n (\sqrt{\frac{x}{a}} + \sqrt{1 + \frac{x}{a}}) \Rightarrow I = a l n (\sqrt{\frac{x}{a}} + \sqrt{1 + \frac{x}{a}})

+ a \frac{\sqrt{x}}{\sqrt{a}} \sqrt{1 + \frac{x}{a}} + C

I = a l n (\frac{\sqrt{x} + \sqrt{a + x}}{\sqrt{a}}) + \sqrt{x} \sqrt{a + x} + C

Commented by mathmax by abdo last updated on 10/Mar/20

I = \int \frac{d x}{s i n x (3 + 2 c o s x)} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

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

= \int \frac{1 + t^{2}}{3 t + 3 t^{3} + 2 t - 2 t^{3}} d t = \int \frac{t^{2} + 1}{t^{3} + 5 t} d t = \int \frac{t^{2} + 1}{t (t^{2} + 5)} d t l e t d e c o m p o s e

F (t) = \frac{t^{2} + 1}{t (t^{2} + 5)} \Rightarrow F (t) = \frac{a}{t} + \frac{b t + c}{t^{2} + 5}

a = t F (t) ∣_{t = 0} = \frac{1}{5}

{l i m}_{t \to + \infty} t F (t) = 1 = a + b \Rightarrow b = 1 - a = \frac{4}{5}

F (1) = \frac{1}{3} = a + \frac{b + c}{6} \Rightarrow 2 = 6 a + b + c \Rightarrow c = 2 - 6 a - b

= 2 - \frac{6}{5} - \frac{4}{5} = 0 \Rightarrow F (t) = \frac{1}{5 t} + \frac{4}{5} \times \frac{t}{t^{2} + 5} \Rightarrow

I = \frac{1}{5} l n ∣ t ∣ + \frac{2}{5} l n (t^{2} + 5) + C

I = \frac{1}{5} l n ∣ t a n (\frac{x}{2}) ∣ + \frac{2}{5} l n (5 + {t a n}^{2} (\frac{x}{2})) + C .

Answered by TANMAY PANACEA last updated on 10/Mar/20

\int \frac{a + x}{\sqrt{x^{2} + a x}} d x

\frac{1}{2} \int \frac{2 x + a + a}{\sqrt{x^{2} + a x}} d x

\frac{1}{2} \int \frac{d (x^{2} + a x)}{\sqrt{x^{2} + a x}} + \frac{a}{2} \int \frac{d x}{\sqrt{{(x + \frac{a}{2})}^{2} - \frac{a^{2}}{4}}}

\frac{1}{2} \times \frac{{(x^{2} + a x)}^{\frac{1}{2}}}{\frac{1}{2}} + \frac{a}{2} l n (x + \frac{a}{2} + \sqrt{{(x + \frac{a}{2})}^{2} - \frac{a^{2}}{4}}

Answered by TANMAY PANACEA last updated on 10/Mar/20

\int \frac{s i n x}{(1 - {c o s}^{2} x) (3 + 2 c o s x)} d x

\int \frac{- d a}{(1 - a^{2}) (3 + 2 a)} [a = c o s x]

\int \frac{d a}{(a + 1) (a - 1) (3 + 2 a)}

\frac{1}{(a + 1) (a - 1) (3 + 2 a)} = \frac{p}{a + 1} + \frac{q}{a - 1} + \frac{r}{3 + 2 a}

1 = p (a - 1) (3 + 2 a) + q (a + 1) (3 + 2 a) + r (a + 1) (a - 1)

p u t a - 1 = 0

1 = q \times 2 \times 5 \to q = \frac{1}{10}

p u t a + 1 = 0

1 = p (- 2) (1) \to p = \frac{- 1}{2}

p u t 3 + 2 a = 0

1 = r (\frac{- 3}{2} + 1) (\frac{- 3}{2} - 1) \to 1 = r (\frac{5}{4}) r = \frac{4}{5}

\int \frac{\frac{- 1}{2}}{a + 1} d a + \int \frac{\frac{1}{10}}{a - 1} d a + \int \frac{\frac{4}{5}}{3 + 2 a} d a

\frac{- 1}{2} l n (a + 1) + \frac{1}{10} l n (a - 1) + \frac{4}{5} \times \frac{1}{2} \int \frac{d a}{a + \frac{3}{2}}

\frac{- 1}{2} l n (c o s x + 1) + \frac{1}{10} l n (c o s x - 1) + \frac{4}{10} l n (c o s x + \frac{3}{2})

Commented by niroj last updated on 10/Mar/20

t h a n k s f o r a l l t o g i v e y o u r b e s t .

Answered by behi83417@gmail.com last updated on 04/Apr/20

\int \sqrt{\frac{a + x}{x}} dx [x = {atg}^{2} t \Rightarrow dx = \frac{2 atgtdt}{\cos^{2} t}]

\Rightarrow I = \int \frac{1}{sint} . \frac{2 atgt}{\cos^{2} t} dt = 2 a \int \frac{dt}{\cos^{3} t} =

= a [sect . tgt + \ln (sect + tgt)] + C =

= a [\sqrt{\frac{x}{a}} . \sqrt{1 + \frac{x}{a}} + \ln (\sqrt{1 + \frac{x}{a}} + \sqrt{\frac{x}{a}})] =

= \sqrt{x^{2} + ax} + a . \ln (\sqrt{a + x} + \sqrt{x}) + C^{'} .

[tgt = \sqrt{\frac{x}{a}}, sect = \sqrt{1 + {tg}^{2} t} = \sqrt{1 + \frac{x}{a}}, C^{'} = C + \frac{a}{2} lna]