+91-705-589-3577
FlagBACK
Flag Integral Calculus> Integration of (1+x-2x 2 ) 1/2 dx Integra...
question mark

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

Ishan jain , 7 Years ago
Grade 12th pass
anser 2 Answers
Susmita
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}
 
 
Last Activity: 7 Years ago
Susmita
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}
Please approve the answer if it helps you.
Last Activity: 7 Years ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Other Related Questions on integral calculus

question mark
The area of a square field is 640000cm2.what is its area in hectares
integral calculus
1 Answer Available

Last Activity: 3 Years ago

question mark
Please solve this integration. I am facing difficulty in it
 How to solve this. Please solve anyone. Humble request
integral calculus
1 Answer Available

Last Activity: 3 Years ago

question mark
Using laplace transform, solve d2y/dt2 + dy/dt = t2 +2t given that y=4 and y’=-2 and t=0
integral calculus
1 Answer Available

Last Activity: 3 Years ago

question mark
If 4th term of an A.p is zero, show that the 8th term is double the 6th term
 
integral calculus
1 Answer Available

Last Activity: 4 Years ago