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Chapter 6: Problem 30

Find the integral. $$ \int x \arcsin x d x $$

Short Answer

Expert verified
\(\int x \arcsin x dx = \frac{1}{2}x^2 \arcsin x - \frac{1}{4}\sqrt{1 - x^2}(2x^2+1) + C\).

Step by step solution

01

Set up the integral

Let us set up the integral using the formula for Integration by Parts. Firstly, set \(u = \arcsin x\), the function that would be easier to differentiate, and the remaining part of the function \(dv = x dx\) which would be easier to integrate.
02

Differentiate and Integrate

Next, differentiate 'u' and integrate 'dv'. When \(u = \arcsin x\), \(du = \frac{1}{\sqrt{1 - x^2}} dx\). And when \(dv = x dx\), \(v = \frac{1}{2} x^2 \).
03

Apply Integration by Parts formula

Substitute 'u', 'v', 'du' and 'dv' back into the formula, \(\int u dv = u v - \int v du\). Thus, we have \(\int x \arcsin x dx = \frac{1}{2}x^2 \arcsin x - \int \frac{1}{2}x^2* \frac{1}{\sqrt{1 - x^2}} dx\).
04

Solve the Integral

Solving the remaining integral will involve applying substitution and standard integration formulae. The final result is \(\frac{1}{2}x^2 \arcsin x - \frac{1}{4}\sqrt{1 - x^2}(2x^2+1) + C\), where 'C' is the constant of integration.

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