# Can indefinite integration be simplified using substitution?

• Crystal037
In summary: First substitution gives: $$\int \frac{1}{\sqrt{x(1-x)}} dx=\int\frac{1}{\sqrt x}\cdot\frac{1}{x} dx$$Then, using the chain rule, you get: $$\int\frac{1}{x}\cdot\frac{1}{\sqrt x}dx=\int\frac{1}{\sqrt x}\cdot\frac{1}{x^2}dx$$Relevant Equations:: Integral of 1/rt(1-x^2)dx = arcsinx
Crystal037
Homework Statement
Integrate 1/(x(1-x))^(1/2)dx
Relevant Equations
Integral of 1/rt(1-x^2)dx = arcsinx
Let x=t^2
Then dx=2t dt
Integral of 1/(x(1-x))^(1/2)dx
= integral of 2tdt/t(1-t^2) ^(1/2)
= integral of 2dt/(1-t^2) ^(1/2)
= 2 arcsin(t) +c
= 2 arcsin(rt(x)) +c.
But the answer in my book is arcsin(2x-1) +c.
Tell me how
2 arcsin(rt(x) +C= arcsin(2x-1) +c
I know the constant will vary for both the answers and both the answers must come equal after some manipulation. Is my answer correct.

Try to prove that ##2 arcsin(x)-arcsin(2x-1)## is a constant.
Just denote be ##y=arcsin x## and then ##2x-1 = 2\sin y -1##, so you need to show that the following is constant: ##2y-arcsin(2\sin y -1)##, if you denote this by ##u## you get: ##2y-u=arcsin(2\sin y -1)## which means that: ##\sin (2y-u) = 2\sin(y)-1##; from here you need to solve a trig equation like in high school.

I meant ##y = arcsin \sqrt{x}##.

Crystal037 said:
Homework Statement:: Integrate 1/(x(1-x))^(1/2)dx
Relevant Equations:: Integral of 1/rt(1-x^2)dx = arcsinx

Let x=t^2
Then dx=2t dt
Integral of 1/(x(1-x))^(1/2)dx
= integral of 2tdt/t(1-t^2) ^(1/2)
= integral of 2dt/(1-t^2) ^(1/2)
= 2 arcsin(t) +c
= 2 arcsin(rt(x)) +c.
But the answer in my book is arcsin(2x-1) +c.
Tell me how
2 arcsin(rt(x) +C= arcsin(2x-1) +c
I know the constant will vary for both the answers and both the answers must come equal after some manipulation. Is my answer correct.
One good thing about doing integration is that you can always check your answer by differentiating it to check you get back the integrand.

I don't understand, however, this variable ##r## that seems to come and go from one line to the next.

PeroK said:
One good thing about doing integration is that you can always check your answer by differentiating it to check you get back the integrand.

I don't understand, however, this variable ##r## that seems to come and go from one line to the next.
There is no variable r I used rt to denote root

JD_PM
PeroK said:
One good thing about doing integration is that you can always check your answer by differentiating it to check you get back the integrand.

I don't understand, however, this variable ##r## that seems to come and go from one line to the next.
Yeah I differentiated it and the answer is correct. Thank you

I am getting LHS as a function of cos2y and sin 2y while RHS only as a function of cos2y. Where am I getting wrong

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Crystal037 said:
I am getting LHS as a function of cos2y and sin 2y while RHS only as a function of cos2y. Where am I getting wrong?
The quickest way must be to differentiate, surely? If the derivative is zero, then the function is a constant. And, setting ##x =0## will give you the constant.

Otherwise, you need the double angle formula.

For integrating, even more simple I think just substitution in original x = t + ½ .

SammyS and archaic
Using a graphing calculator you can find that ##\displaystyle 2\arcsin(\sqrt{x}\ ) - \arcsin(2x-1) ## is ##\dfrac{\pi}{2} ## .

This suggests that ##\displaystyle 2\arcsin(\sqrt{x}\ ) - \dfrac{\pi}{2} = \arcsin(2x-1) ##.

Take the sine of the left hand side and you get ##2x-1##, after a bit of manipulation.

$$\int \frac{1}{\sqrt{x(1-x)}} dx=\int\frac{1}{\sqrt x}\cdot\frac{1}{\sqrt{1-x}} dx$$
Doesn't ##\frac{1}{\sqrt x}## remind you of the derivative of some function? Don't you think that ##x=(\sqrt x)^2##?

epenguin said:
For integrating, even more simple I think just substitution in original x = t + ½ .
Nice one!

## What is indefinite integration?

Indefinite integration is the process of finding an antiderivative or primitive function of a given function. It is the reverse operation of differentiation and is used to find the original function when its derivative is known.

## What is the difference between indefinite and definite integration?

Indefinite integration deals with finding an antiderivative while definite integration involves finding the area under a curve between two given points. Indefinite integration results in a family of functions while definite integration gives a single numerical value.

## What are the basic rules of indefinite integration?

The basic rules of indefinite integration include the power rule, constant multiple rule, sum and difference rule, and substitution rule. These rules allow us to find the antiderivative of polynomials, exponential, logarithmic, and trigonometric functions.

## How is indefinite integration used in real-world applications?

Indefinite integration is used in various fields such as physics, engineering, economics, and statistics to model and solve real-world problems. It is used to calculate areas, volumes, and displacements, and to find the velocity and acceleration of objects.

## What are some common techniques used in indefinite integration?

Some common techniques used in indefinite integration include u-substitution, integration by parts, trigonometric substitution, partial fractions, and trigonometric identities. These techniques are used to simplify the given function and make it easier to find its antiderivative.

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