Evaluate
∫ 1 arcsin ( 2 x ) 1 − 4 x 2 d x \displaystyle \int\frac{{1}}{{{\arcsin{{\left({2}{x}\right)}}}\sqrt{{{1}-{4}{x}^{{2}}}}}}{\left.{d}{x}\right.} ∫ arcsin ( 2 x ) 1 − 4 x 2 1 d x for
0 < x < 1 2 \displaystyle {0}<{x}<\frac{{1}}{{2}} 0 < x < 2 1 .
First make the substitution u =
Preview Question 6 Part 1 of 4
Then
∫ 1 arcsin ( 2 x ) 1 − 4 x 2 d x = ∫ \displaystyle \int\frac{{1}}{{{\arcsin{{\left({2}{x}\right)}}}\sqrt{{{1}-{4}{x}^{{2}}}}}}{\left.{d}{x}\right.}=\int ∫ arcsin ( 2 x ) 1 − 4 x 2 1 d x = ∫ Preview Question 6 Part 2 of 4
du
Now integrate with respect to u to get
Preview Question 6 Part 3 of 4
+ C
So
∫ 1 arcsin ( 2 x ) 1 − 4 x 2 d x \displaystyle \int\frac{{1}}{{{\arcsin{{\left({2}{x}\right)}}}\sqrt{{{1}-{4}{x}^{{2}}}}}}{\left.{d}{x}\right.} ∫ arcsin ( 2 x ) 1 − 4 x 2 1 d x =
Preview Question 6 Part 4 of 4
+ C
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Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c