Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to find the integral $$\int \sin^3xdx\\= \int \sin^2x \sin xdx\\= \int (1-\cos^2x) \sin xdx$$

Substitution: $$z=\cos x$$ $$\frac{dz}{dx} = -\sin x$$ $$-dz = \sin x dx$$

Now the above expression would be like this $$\int -(1-z^2) dz$$ Now integration would be $$-z + \frac{z^3}{3} + c$$ we replace $z$ by $\cos x$ so our answer would be $$-\cos x + \frac{\cos^3x}{3} + c$$

But in book this answer is not correct. I want to know the error. Please, can any one solve it and tell me about the error?

share|improve this question
1  
Please format your question using $\LaTeX$ enclosed in dollar signs. –  Your Ad Here Jan 27 at 10:43
1  
I'm sorry to ask you such a personal question zoonie, but how old are you? –  Tomáš Zato Jan 27 at 10:48
1  
@zonnie, it isn't nice to ask a further question in the comments without even addressing other comments... –  DonAntonio Jan 27 at 10:50
3  
Your answer is correct; there is no error. Nice solution, in fact. The answer in the book may have simplified $-\cos x +(\cos^3 x)/3$ further. –  David Mitra Jan 27 at 11:14
1  
@zonnie. May be, you could answer the questions you have been asked. –  Claude Leibovici Jan 27 at 11:30

1 Answer 1

This method looks easier.You can use $sin3x$=3$sinx$-4$sin^3x$. Hence you will get $sin^3x$=$\frac{3sinx-sin3x}{4}$.Hence $$\int \sin^3(x)dx\\=\int \frac{3sinx-sin3x}{4}= \int \frac{3sinx}{4}-\int \frac{sin3x}{4}=-\frac{3cosx}{4}+\frac{cos3x}{12}+c$$

share|improve this answer
    
Why $\sin$ instead of $\cos$? –  anorton Jan 27 at 12:55
    
Its not like converting sin to cos.This equation i used to convert the third power to unity power,hence only in terms of sin. –  Devgeet Patel Jan 27 at 12:58
    
Oh! sorry--I originally thought this was a hint simplify the base. That's an awesome identity. :) (+1) –  anorton Jan 27 at 23:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.