Question

anyone can help me to evaluate this integrals

Answer 1

\[\int\limits_{}^{}(1-x^2)^3\]

Answer 2

i just want to clarify,,,is it \[\frac{ (1-x^2)^4 }{ 4 }+c\]

Answer 3

no, you are not just applying the power rule to that

Answer 4

although you are thinking in the right direction.

Answer 5

hmm how can we do this? ^^

Answer 6

\(\color{#000000}{\displaystyle \frac{d}{dx} \left[\frac{1}{4}(1-x^2)^4\right]=\frac{1}{4}\cdot 4(1-x^2)^{4-1} \color{blue}{\times {\rm chain~rule}}}\)

Answer 7

What chain rule completes the derivative?

Answer 8

i dont really know this one

Answer 9

you need the chain rule for 1-x², right?

Answer 10

woah! i feel dizzy all of sudden

Answer 11

I feel that you need to expand it, if you don't know about trig-substitution

Answer 12

\(\color{#000000}{\displaystyle (a-b)^3=a^3-3a^2b+3ab^2-b^3}\)

and in your case,
\(\color{#000000}{\displaystyle a=1}\)
\(\color{#000000}{\displaystyle b=x^2}\)

and so you get,
\(\color{#000000}{\displaystyle (1-x^2)^3=(1)^3-3(1)^2(x^2)+3(1)(x^2)^2-(x^2)^3}\)

Answer 13

\(\color{#000000}{\displaystyle (1-x^2)^3=1-3x^2+3x^4-x^6}\)

Answer 14

\(\color{#000000}{\displaystyle \int (1-x^2)^3~dx= \int (1-3x^2+3x^4-x^6)~dx}\)

Answer 15

\[=x-x^2+3/5 x^5 -1/7 x^7 +c?\]

Answer 16

And then apply the rules.

\(\color{#000000}{\displaystyle \int x^n~dx= \frac{1}{n+1}x^{n+1}\color{grey}{\rm +C}}\)

\(\color{#000000}{\displaystyle \int a~dx= ax\color{grey}{\rm +C}}\)

Answer 17

Some of the terms are incorrect

Answer 18

one term is wrong

Answer 19

-x^3 sorry lol

Answer 20

yes, right

Answer 21

really, this was very easy,but can you teach me how to apply trig sub here in this problem ^^

Answer 22

\(\color{#000000}{\displaystyle =x-x^3+(3/5)x^5-(1/7)x^7+C}\)

Answer 23

I have to go for an hour right now, but I am for sure online at 7:00 pm.
If you wish to learn u-sub and trig-sub, you can look it up on lamar tutorial.

Everything is well written out, and is easily readable.
(Not too scholarly at all)

Answer 24

just google «lamar tutorial u-substitution»

Answer 25

But, for this problem particularly, the trig sub would be

x=sinθ
dx=cosθ dθ

Answer 26

wait let me ask this problem \[\int\limits_{}^{}(4x+1)^2dx\]

Answer 27

igtg. bye

Answer 28

you can do u=4x+1, if you wish to do u sub

Answer 29

is this the same process of what we done?

Answer 30

i reallly got to go. Sorry

Answer 31

Ok, I am back!

Answer 32

Chain rule can definitely be proven, but
for this purpose will take use it as given. \(\tiny \\[0.8em]\)
Let's proceed! \(\tiny \\[0.8em]\)
By the chain rule you know that: \(\tiny \\[0.8em]\)
\(\color{#000000 }{ \displaystyle \frac{d}{dx}~{\rm H}\left({\rm f}(x)\right)={\rm H}{\tt{\tiny~} '}\left({\rm f}(x)\right) \cdot {\rm f}{\tt{\tiny~} '}(x) }\)

And therefore, \(\tiny \\[0.8em]\)
\(\color{#000000 }{ \displaystyle \int {\rm H}{\tt{\tiny~} '}\left({\rm f}(x)\right) \cdot {\rm f}{\tt{\tiny~} '}(x)~{\mathrm d}x={\rm H}\left({\rm f}(x)\right)\color{#666666}{\rm +C} }\) \(\tiny \\[1.3em]\)
This is the integration-differentiation relationship.
And here, actually u-substitution comes into the
play! \(\tiny \\[0.8em]\)
\(\color{#ff0000 }{ \displaystyle \sf So,{\tiny~~~}suppose{\tiny~~~}that{\tiny~~~}I{\tiny~~~}want{\tiny~~~}to{\tiny~~~}solve{\tiny~~~}this{\tiny~~~}integral, }\) \(\tiny \\[0.7em]\)
\(\color{#000000 }{ \displaystyle \int {\rm H}{\tt{\tiny~} '}\left({\rm f}(x)\right) \cdot {\rm f}{\tt{\tiny~} '}(x)~{\mathrm d}x }\)
and I'd like to do this is through \(\color{#ff0000 }{ \displaystyle \sf u-substitution }\)!
(not by recognizing derivatives) \(\tiny \\[0.7em]\)
For this purpose I will set,
\(\color{#000000 }{ \displaystyle {\sf u}={\rm f}(x) }\) \(\tiny \\[0.7em]\)

you also know that will help you find an
the derivative equivalent differential for
of f(x), so ... \(\color{ }{ {\sf u} }\). ((both sides times \(\color{ }{ d{\sf u} }\))) \(\tiny \\[0.9em]\)
\(\color{#000000 }{ \displaystyle \frac{d{\sf u}}{dx} ={\rm f}{\tt {\tiny~}'}(x) \quad \Longrightarrow \quad d{\sf u}=f'(x)~dx}\)

The old integral, becomes, \(\tiny \\[0.6em]\)
\(\color{#000000 }{ \displaystyle \int {\rm H}{\tt{\tiny~} '}\left(\color{blue}{{\rm f}(x)}\right) \color{red}{{\rm f}{\tt{\tiny~} '}(x)~{\mathrm d}x} = \int {\rm H}{\tt{\tiny~} '}\left(\color{blue}{{\sf u}}\right) \color{red}{{\mathrm d}{\sf u}} }\)

And then you get:
\(\color{#000000 }{ \displaystyle \int {\rm H}{\tt{\tiny~} '}\left({\sf u}\right) {\mathrm d}{\sf u}= {\rm H}\left({\sf u}\right)\color{#666666}{\rm +C} }\)

Then, substitute the original variable back, you get:
\(\color{#000000 }{ \displaystyle {\rm H}\left(f(x)\right)\color{#666666}{\rm +C} }\) (as the final answer)

Answer 33

An example problem, similar to yours,
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \left(6x+5\right)^{10}~dx}\)

the substitution is as follows:
\(\color{#000000}{\displaystyle u= 6x+5 }\)
\(\color{#000000}{\displaystyle du/dx= 6\quad \Rightarrow \quad du=6\cdot dx\quad \Rightarrow \quad \frac{1}{6}du=dx }\)

So the old integral, becomes, \(\tiny\\[0.8em]\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \left(6x+5\right)^{10}~dx\quad \Longrightarrow \Longrightarrow\quad \int\limits_{~}^{~}(u)^{10}\left(\frac{1}{6}du\right) }\)

And then, this becomes,\(\tiny\\[0.8em]\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~}(u)^{10}\left(\frac{1}{6}du\right)= \frac{1}{6}\int\limits_{~}^{~}u^{10}du}\)\(\tiny\\[1.2em]\)
\(\color{#000000}{\displaystyle=\frac{1}{6}\frac{u^{10+1}}{10+1}\color{#666666}{\rm +C}=\frac{1}{66}u^{11}\color{#666666}{\rm +C}=\frac{1}{66}(6x+5)^{11}\color{#666666}{\rm +C}}\)

Answer 34

And in general, for \(\small m\ne -1\),

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \left(ax+b\right)^{m}~dx}\)

the substitution is as follows:

\(\color{#000000}{\displaystyle u= ax+b }\)
\(\color{#000000}{\displaystyle du/dx= a\quad \Rightarrow \quad du=a\cdot dx\quad \Rightarrow \quad \frac{1}{a}du=dx }\)

So the old integral, becomes, \(\tiny\\[0.8em]\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \left(ax+b\right)^{m}~dx\quad \Longrightarrow \Longrightarrow\quad \int\limits_{~}^{~}(u)^{m}\left(\frac{1}{a}du\right) }\)

And then, this becomes,\(\tiny\\[0.8em]\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~}(u)^{10}\left(\frac{1}{a}du\right)= \frac{1}{a}\int\limits_{~}^{~}u^{10}du}\)\(\tiny\\[1.2em]\)
\(\color{#000000}{\displaystyle=\frac{1}{a}\frac{u^{m+1}}{m+1}\color{#666666}{\rm +C}=\frac{1}{a(m+1) }u^{m+1}\color{#666666}{\rm +C}=\frac{(ax+b)^{m+1}}{am+a }\color{#666666}{\rm +C}}\)

So you can conclude that
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \left(ax+b\right)^{m}~dx=\frac{(ax+b)^{m+1}}{am+a }\color{#666666}{\rm +C}}\)

Answer 35

\(\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{1-x^2}}~dx }\)

Trig substitution,
\(\color{#000000 }{ \displaystyle x=\sin \theta }\)
\(\color{#000000 }{ \displaystyle dx/d\theta=\cos\theta \quad \Longrightarrow \quad dx= \cos \theta ~d\theta }\)

So, your integral becomes,
\(\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{1-\left[\sin \theta\right]^2}}~\left[\cos \theta ~d\theta \right] }\)

And when you simplify this, you get,
\(\color{#000000 }{ \displaystyle \int \frac{\cos \theta}{\sqrt{1-\sin^2 \theta}} ~d\theta }\)

\(\color{#000000 }{ \displaystyle \int \frac{\cos \theta}{\sqrt{\cos^2 \theta}} ~d\theta }\)

and this simplifies to,

\(\color{#000000 }{ \displaystyle \int 1~d\theta =\theta \color{#666666}{\rm +C} }\)

Then, we need to solve for \(\theta\) in terms of \(x\).
Our initial substitution was \(\color{#000000 }{ \displaystyle x=\sin \theta }\), and
therefore,

\(\color{#000000 }{ \displaystyle x=\sin \theta\quad\quad \Longleftrightarrow \quad\quad\theta =\sin^{-1}x }\)

And therefore, we get that:
\(\color{#000000 }{ \displaystyle \int \frac{1}{\sqrt{1-x^2}}~dx=\sin^{-1}x \color{#666666}{\rm +C} }\)