Question-40675

Question Number 40675 by Raj Singh last updated on 26/Jul/18

Commented by math khazana by abdo last updated on 30/Jul/18

l e t I = \int \frac{d x}{2 s i n x + c o s x + 3} c h a 7 g e m e n t

t a n (\frac{x}{2}) = t g i v e

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

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

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

=_{t + 1 = t a n θ} \int \frac{(1 + {t a n}^{2} θ) d θ}{1 + {t a n}^{2} θ} = θ + c

= a r c t a n (1 + t) + c

= a r c t a n (1 + t a n (\frac{x}{2})) + c .

Commented by math khazana by abdo last updated on 30/Jul/18

s i r R a j i t h i n k t h e Q i s p r o v e n o t s o l v e \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

\int \frac{d x}{2 s i n x + c o s x + 3}

l e t t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x

\frac{2 d t}{1 + t^{2}} = d x

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

\int \frac{2 d t}{4 t + 1 - t^{2} + 3 + 3 t^{2}}

\int \frac{2 d t}{2 t^{2} + 4 t + 4}

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

{t a n}^{- 1} (\frac{t + 1}{1}) + c

{t a n}^{- 1} (\frac{1 + t a n \frac{x}{2}}{1}) + c

Commented by Raj Singh last updated on 26/Jul/18

t t t h h a n n k s

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

i t s o k \dots