sin-1-2x-2-4x-2-8x-13-dx-

Question Number 84556 by jagoll last updated on 14/Mar/20

\int \sin^{- 1} (\frac{2 x + 2}{\sqrt{{4 x}^{2} + 8 x + 13}}) dx

Commented by john santu last updated on 14/Mar/20

\int \sin^{- 1} (\frac{2 x + 2}{\sqrt{{(2 x + 2)}^{2} + 9}}) dx

let 2 x + 2 = 3 \tan θ \Rightarrow 2 dx = 3 \sec^{2} θ d θ

\int \sin^{- 1} (\frac{3 \tan θ}{3 \sec θ}) \times \frac{3}{2} \sec^{2} θ d θ =

\int \frac{3}{2} \sin^{- 1} (\sin θ) d (\tan θ) =

\frac{3}{2} \int θ d (\tan θ) = \frac{3}{2} [θ \tan θ - \int \tan θ d θ]

= \frac{3}{2} [θ \tan θ + \int \frac{d (\cos θ)}{\cos θ}]

= \frac{3}{2} [θ \tan θ + \ln ∣ \cos θ ∣] + c

= \frac{3}{2} [(\frac{2 x + 2}{3}) \tan^{- 1} (\frac{2 x + 2}{2}) + \ln ∣ \frac{3}{\sqrt{{4 x}^{2} + 8 x + 13}} ∣] + c

Answered by mind is power last updated on 14/Mar/20

= \int {s i n}^{-} (\frac{2 x + 2}{\sqrt{{(2 x + 2)}^{2} + 3^{2}}}) = \int {s i n}^{-} (\frac{\frac{2 x + 2}{3}}{\sqrt{1 + {(\frac{2 x + 2}{3})}^{2}}})

{s i n}^{-} (\frac{r}{\sqrt{r^{2} + 1}})

r = t g (θ) \Rightarrow {s i n}^{-} (\frac{t g (θ)}{\sqrt{1 + {t g}^{2} (θ)}}) = {s i n}^{-} (s i n (θ)) = θ = a r c t a n (r)

\Rightarrow \int {s i n}^{-} (\frac{\frac{2 x + 2}{3}}{\sqrt{1 + {(\frac{2 x + 2}{3})}^{2}}}) d x = \int a r c t a n (\frac{2 x + 2}{3}) d x b y p a r t e a s y n o w

Commented by jagoll last updated on 14/Mar/20

thank you sir . i can try

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