Evaluate the following integral: \int \frac{1-x}{\sqrt{1-x^{2}}}dx

lunnatican4

lunnatican4

Answered question

2022-01-05

Evaluate the following integral:
1x1x2dx

Answer & Explanation

Papilys3q

Papilys3q

Beginner2022-01-06Added 34 answers

Step 1: Given that
Evaluate the following integral:
1x1x2dx
Step 2: Solve
We have,
1x1x2dx=11x2dxx1x2dx
=sin1(x)xt×dt2x
=sin1x+12[t12+112+1]+C
=sin1x+t+C
=sin1x+1x2+C
Robert Pina

Robert Pina

Beginner2022-01-07Added 42 answers

Given:
1x1x2dx
11x2x1x2dx
Use properties
11x2dxx1x2dx
arcsin(x)+1x2
Solution:
arcsin(x)+1x2+C
karton

karton

Expert2022-01-11Added 613 answers

1x1x2dx=(11x2x1x2)dx=11x2dxx1x2dxNow calculate11x2dx=arcsin(x)Now calculatex1x2dx=121udu1udu=2u121udu=u=1x211x2dxx1x2dx=arcsin(x)+1x2=arcsin(x)+1x2+C

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