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Integration of (1+x-2x 2 ) 1/2 dx Integration of sininversex/(1-x 2 ) 3/2

  1. Integration of (1+x-2x2)1/2dx
  2. Integration of sininversex/(1-x2)3/2

Grade:12th pass

2 Answers

Susmita
425 Points
3 years ago
Look at the denominator term within root.Try to write it as two whole square terms.
1+x-2x2
=1-2(x2-x/2)
=1-2(x^2-2.x.\frac{1}{4}+\frac{1}{16})+\frac{1}{8}
=\frac{9}{8}-2(x-\frac{1}{4})^2
=2[(\frac{3}{4})^2-(x-\frac{1}{4})^2]
So the integration becomes
\frac{1}{2}\int \frac{dx}{\sqrt{(3/4)^2-(x-1/4)^2}}
=\frac{1}{\sqrt{2}}sin^-1\frac{x-1/4}{3/4}
 
 
Susmita
425 Points
3 years ago
In the 2nd integration put
z=sin-1x
So, dz=\frac{dx}{\sqrt{1-x^2}}
Also x=sinz
put these values in the integration.It will become
\int\frac{zdz}{cos^2z}
=\int\ z sec^2z dz
Now do integration by parts.Take z as 1st function.
\int\ z sec^2z dz
 
=ztanz-\int tanz dz
=ztanz+logcoz
=\frac{xsin^-1x}{\sqrt{1-x^2}}+log\sqrt{1-x^2}
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