Question

Kindly help with this integration

Answer 1

\[\int\limits_{0}^{\pi}(1-\cos^2\theta)\sin\theta d\theta\]

Answer 2

\[\rm= \int_{0}^{\pi}\sin^2 \theta d\theta\]

Answer 3

\[\int {\pi to 0} sin^3 \theta d\theta \]

Answer 4

I meant : integration (pi to 0) sin^3 theta d theta.

Answer 5

The next step is

\[\int\limits_{0}^{\pi} -(1-cos^2\theta)d(cos\theta)\]

Answer 6

How do we get this step ?

Answer 7

Sorry, I got a little confused.\[\int_{0}^{\pi} \sin^3 \theta d\theta\]

Answer 8

I got that Parth but that step won't help in the derivation I want to do.

Tell me how do we get the next step I mentioned above

Answer 9

is that next step even right or not?

Answer 10

That step seems to be right because that gives the correct answer

Answer 11

\[\sin^3 \theta = - \cos \theta \sin \theta \] it doesn't seem to me.

Answer 12

does it give the answer correct with : \[\sin^3 \theta \] ?

Answer 13

Remember the d(cos theta)

\[d\theta -----> d(cos\theta)\]

Answer 14

It gives the correct answer but that would increase the number of steps.... and of course much more time to do....

Answer 15

@experimentX

Answer 16

@calculusfunctions

Answer 17

OK what can I help you with?

Answer 18

hmm i don't get the step u have done or how u get that.. however i have a different approach in my mind to solve the question i can if u want.

Answer 19

sir please help me with the integration

Answer 20

Have you tried the substitution method?

Answer 21

That's where I am stuck

What do I have to substitute

Answer 22

Sorry the better question is, do you know the substitution method?

Answer 23

I have to get this

\[\int\limits_{0}^{\pi}(1-\cos^2\theta)\sin\theta d\theta\]

to this step

\[\int\limits_{0}^{\pi} -(1-cos^2\theta)d(cos\theta)\]

Answer 24

No, That is not a good way to start this problem.

Answer 25

I know that sir

But where do I have to use substitution here and how ?

Answer 26

What do you think you should let u equal? Think carefully.

Answer 27

@Zekarias please do not give answers.

Answer 28

\[\sin(\theta)d \theta=-\frac{d(\cos (\theta))}{d \theta}d \theta=-d(\cos( \theta))\]

Answer 29

Just the answer: http://www.wolframalpha.com/input/?i=integrate+from+0+to+pi+%281-cos%5E2+x%29+sin+x+dx

http://www.wolframalpha.com/input/?i=integrate+%281-cos%5E2+x%29+sin+x+dx

Answer 30

Sir I couldn't get you question

Answer 31

@vishweshshrimali5 you asked for my help so if you'd like for me to teach you thaen please ask people to stop giving out answers.

Answer 32

@everyone kindly pllease wait for sometime

Answer 33

Sir you may continue.... Please

Answer 34

Thank you! Now\[\int\limits_{0}^{\pi}(1-\cos ^{2}\theta)\sin \theta d \theta\]is what we have correct?

Answer 35

Yes sir

Answer 36

OK so tell me what are the derivatives of sine and cosine function.

Answer 37

u means

\[\cfrac{d}{dx} (sin x) = cos x\]

and
\[\cfrac{d}{dx} (cos x) = -sin x\]

Answer 38

Correct! so then u = ? and du = ? What do you think?

Answer 39

u = cos x

and du = -sin x dx

Answer 40

Keep in mind that sin θdθis outside the parentheses when you make your decision.

Answer 41

yep

Answer 42

Yes! Perfect!! Now what do you think?

Answer 43

I think

lets take \[sin\theta d\theta = -d(cos\theta)\]

Answer 44

Correct ?

Answer 45

No, go ahead and substitute but you now also have to change the intervals in terms of u as well. Do you know what I mean by that?

Answer 46

Yes

u mean now the intervals will be for cos theta

Answer 47

\[\int\limits_{0}^{\pi}(1-\cos ^{2}\theta)\sin \theta d \theta \]
\[=- \int\limits_{1}^{-1}(1-u ^{2})du\]

Do you understand what I did?

Answer 48

Yep

Answer 49

And now I should put
u = cos theta

Answer 50

and integrate it

Answer 51

\[\int\limits_{b}^{a}f(x)dx =- \int\limits_{a}^{b}f(x)dx\]

This is a theorem. Do you know this?

Answer 52

Yes

Answer 53

Interchanging the limits and adding a -ve sign

Answer 54

So then we have\[\int\limits_{-1}^{1}(1-u ^{2})du\]
Correct?

Answer 55

Yes

Answer 56

Now evaluate this integral. Show me your work please!

Answer 57

Wait!

Answer 58

First ask yourself, is\[1-u ^{2}\]an even function?

Answer 59

Ok

\[\int\limits_{-1}^{1}(1-u ^{2})du\]

\[= \int\limits_{-1}^{1}du - \int\limits_{-1}^{1}(u^2 du)\]

\[= 1-(-1) - \cfrac{1}{3}(1-(-1))\]
\[= 2 - \cfrac{2}{3}\]
\[=\cfrac{4}{3}\]

Answer 60

\[f(u) = 1-u^2\]

\[f(-u) = 1-(-u)^2 = 1-u^2 = f(u)\]

Even function

Answer 61

Am I correct sir ?

Answer 62

If f(x) is an even function, then\[\int\limits_{-a}^{a}f(x)dx =2\int\limits_{0}^{a}f(x)dx\]Are you aware of this theorem?

Answer 63

Yes I could have used it too

Answer 64

It doesn't matter now because you got already got the right answer but you should always be conscious of the theorems since they usually make life easier. Alright?

Answer 65

Yes sir

I will remember it

Thanks a lot for your help

You really are a wonderful teacher

Answer 66

Excellent work @vishweshshrimali5 you're obviously very bright.

Answer 67

Thanks a lot sir

Answer 68

@vishweshshrimali5 thanks! Now before I sign off, do you need anything else right now?

Answer 69

You're welcome!

Answer 70

No sir

Thanks a lot again

Answer 71

Alright then Bye! If you need me the next time I sign in, let me know like you did today. Take Care!

Answer 72

Bye sir

Answer 73

Thanks everyone else for your help