Evaluate the following integrals. ∫012sin−1x1−x2dx

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Evaluate the following integrals.  ∫012sin−1x1−x2dx

soanooooo40 · Accepted Answer

Step 1 Given: I=∫012sin−1x1−x2dx...(1) for evaluating given integral we substitute sin−1⁡x=t...(2) if, x=0 then t=0 and if, x=12 then t=π6 now differentiating equation (2) with respect to x so, ddx(sin−1x)=ddx(t)   (∵ddx(sin−1x)=11−x2) 11−x2=dtdx dx1−x2=dt Step 2 now replacing sin−1x with t, dx1−x2 with dt and changing limits also in equation (1) so, ∫012sin−1xdx1−x2=∫0π6tdt   (∵∫tdt=t22+c) =[t22]0π6 =12[(π6)2−(0)2] =12(π236) =π272 hence, given integral is equal to π272.

Dabanka4v · Accepted Answer

Make the substitution, and find the corresponding limits in terms of u.  u=sin−1x   du=11−x2dx  a=sin−10=0  b=sin−112=π6  ∫012sin−1x1−x2dx=∫0π6udu  =[12u2]0π6  =12[π236−0]  =π272  Result:  π272

Vasquez · Accepted Answer

We need to evaluate the definite integral ∫012sin−1⁡x1−x2dx We will use substitution. Substitute u=sin−1⁡x⇒du=11−x2dx; ⇒∫012udu Integrate by applying ∫xn=xn+1n+1; ⇒u22 Revert the substitution and evaluate; ⇒(sin−1⁡x)22|012=(sin−1⁡12)22−(sin−1⁡0)22 =(π6)22−022 =π2362 =π272 Result: ∫012sin−1⁡x1−x2dx=π272

Evaluate the following integrals. \int_{0}^{1/2}\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}dx

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