Integrate the function- -sin -12 x -1- x 2 dx-

Let I =∫ sin^ 1 ( 2x/1+x^2 )dx On putting x =tan θ, we get I =∫ sin^ 1 ( 2 tan θ/1+tan^2 θ )dx =∫ sin^ 1(sin 2 θ)dx ( ∵ sin 2 θ =2 tan θ/1+tan^2 θ ) =2 ∫θ dx ...

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Byju's Answer

Standard XII

Mathematics

Theorems on Integration

Integrate the...

Question

Integrate the function.
$\int s i n^{- 1} (\frac{2 x}{1 + x^{2}}) d x .$

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Solution

Let $I = \int s i n^{- 1} (\frac{2 x}{1 + x^{2}}) d x$
On putting $x = t a n θ$ , we get $I = \int s i n^{- 1} (\frac{2 t a n θ}{1 + t a n^{2} θ}) d x$
$= \int s i n^{- 1} (s i n 2 θ) d x (∵ s i n 2 θ = \frac{2 t a n θ}{1 + t a n^{2} θ}) = 2 \int θ d x = 2 \int t a n^{- 1} x d x [∵ x = t a n θ \Rightarrow θ = t a n^{- 1} x] = 2 \int 1 I I . t a n^{- 1} I x d x = 2 t a n^{- 1} x \int 1 d x - 2 \int [\frac{d}{d x} t a n^{- 1} x \int 1 d x] d x$
(using Integration by parts)
$\Rightarrow I = 2 t a n^{- 1} x . x - 2 \int [\frac{1}{1 + x^{2}} . x] d x$
Put $1 + x^{2} = t \Rightarrow 2 x = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{2 x}$
$∴ I = 2 x t a n^{- 1} x - 2 \int [\frac{x}{t} . \frac{d t}{2 x}] = 2 x t a n^{- 1} x - \int \frac{1}{t} d t = 2 x t a n^{- 1} x - l o g | t | + C \Rightarrow I = 2 x t a n^{- 1} x - l o g | 1 + x^{2} | + C$

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