Question

integration of x sin inverse x

Answer 1

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Answer 2

Answer:

$\frac{x^2}{2}sin^{-1}x - \frac{1}{4}sin^{-1}x + \frac{x\times \sqrt(1 -x^2)}{4} + C$

Step-by-step explanation:

According to this question

Given that

$\int x sin^(-1)x dx$

$sin^{-1}x\int x dx - \int [\frac{d}{dx}(sin^{-1}) \int x dx] dx$

$sin^{-1}x\times \frac{x^2}{2} - \int [\frac{1}{\sqrt{1 - x^2}} \frac{1}{2}\times x^2] dx$

$sin^{-1}x\times \frac{x^2}{2} - \frac{1}{2}\int [\frac{ x^2}{\sqrt{1 - x^2}}] dx$

$sin^{-1}x\times \frac{x^2}{2} + \frac{1}{2}\int [\frac{ 1 -x^2 - 1}{\sqrt{1 - x^2}}] dx$

$sin^{-1}x\times \frac{x^2}{2} + \frac{1}{2}\int [\frac{ 1 -x^2}{\sqrt{1 - x^2}} - \int \frac{1}{\sqrt{1 - x^2}}] dx$

$sin^{-1}x\times \frac{x^2}{2} + \frac{1}{2}\int [{\sqrt{1 - x^2}} - \int \frac{1}{\sqrt{1 - x^2}}] dx$

$sin^{-1}x\times \frac{x^2}{2} + \frac{1}{2}[\frac{x\times \sqrt(1 -x^2)}{2} +  \frac{1}{2}sin^{-1}x - sin^{-1}x] + C$

$sin^{-1}x\times \frac{x^2}{2} + \frac{1}{2}[\frac{x\times \sqrt(1 -x^2)}{2} -  \frac{1}{2}sin^{-1}x]$

$\frac{x^2}{2}sin^{-1}x - \frac{1}{4}sin^{-1}x + \frac{x\times \sqrt(1 -x^2)}{4} + C$