Question

Please Solve, I can't really reach the solution..​

Answer 1

$\sf\huge{\underline{Question :}}$

$\int\limits { \tan(\arcsin(x)} \, dx$

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$\sf\huge{\underline{Solution :}}$

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Tan θ = (Sin θ)/ (Cos θ)

from, Sin²θ + Cos²θ = 1 identity we know that,

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Cos θ =

$\sqrt{1 - { \sin^{2}(\theta) } }$

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Puting this in Tan θ we get,

$\tan(\theta) = \dfrac{ \sin(\theta) }{\sqrt{1 - { \sin^{2}(θ) } } }$

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Now, arcsin x = θ

then Sin θ = x

so,putting this it becomes :

$\tan(\arcsin(x)) = \dfrac{ x }{ \sqrt{1 - {x}^{2} } }$

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Now, we put this instead of Tan(arcsin) in our initial eq.

$\int\limits { \dfrac{ x }{ \sqrt{1 - {x}^{2} } }} \, dx$

let 1-x² = t

dt/dx = -2x

dt/-2x = dx

$\int\limits { \dfrac{ x }{ \sqrt{t} }} \, \dfrac{dt}{ - 2x}$

-1/2 comes out — constant

x get cut {divided}

$\int\limits { \dfrac{1}{ \sqrt{t} } } \, dt$ is left

t^-½ so, using power rule

it becomes $\dfrac{ 2 \sqrt{t} }{1}$

outside it's -1/2

so it becomes -√t

puting value of t we get,

$- \sqrt{1 - {x }^{2} } \: \: + C$

Answer 2

Hey There

Here's The Answer

________________________

Kindly refer to the Attachment.

Hope It Helps