help@askiitians.com

+91-87964 74404

Grade 12th passIntegral Calculus

Integration of (1+x-2x²)^1/2dx

Integration of sininversex/(1-x²)^3/2

Ishan jain

7 Years agoGrade 12th pass

2 Answers

Susmita

7 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

7 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.

$\int\ z sec^2z dz$

$=ztanz-\int tanz dz$

$=ztanz+logcoz$

$=\frac{xsin^-1x}{\sqrt{1-x^2}}+log\sqrt{1-x^2}$

Please approve the answer if it helps you.

The area of a square field is 640000cm².what is its area in hectares

integral calculus

1 Answer Available

Last Activity: 3 Years ago

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: 4 Years ago

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: 4 Years ago

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

View all Questions