Textbook says the integral of (sin cos^3) is
-(cos^4)/4 . I'm using trig integrals technique which instead gives me (sin^2)/2 - (sin^4)/4.
What gives?

Question

Textbook says the integral of (sin cos^3) is 
-(cos^4)/4 .  I'm using trig integrals technique which instead gives me (sin^2)/2  -  (sin^4)/4.
What gives?

rational · Answer

$$\large \int \sin x \,\cos^3x \, dx$$

rational · Answer

like that ?

anonymous · Answer

Yes

rational · Answer

what technique are you using ? could you elaborate

anonymous · Answer

Sure, I turned cos^3 into cos cos^2.  Then turned cos^2 into (1-sin^2).  u-subbed sin and integrated to the above answer

rational · Answer

Looks legit! so with your method, you get something like this ?$$\int\sin x~\cos x~(\cos^2x)\, dx =\int\sin x~\cos x~(1-\sin^2x)\, dx= \int u(1-u^2)\, du$$

anonymous · Answer

exactly

rational · Answer

$$ \int u(1-u^2)\, du=\frac{u^2}{2}-\frac{u^4}{4}+C = \frac{\sin^2x}{2}-\frac{\sin^4x}{4}+C$$

rational · Answer

you want to match your answer wid textbook is it ?

rational · Answer

btw, your answer is also correct

anonymous · Answer

Yea, i plugged in some angles to see if maybe it would result in the same answer but no dice.   They still don't match :(

rational · Answer

they match upto a " constant "

rational · Answer

let me ask you a quick question, what are the derivatives of below two functions ? 

1)  $f(x)=\sin x$

2)  $g(x) = \sin x+100000$

anonymous · Answer

both are cosx

rational · Answer

so whats "the" antiderivative of $\cos x$ ?

anonymous · Answer

sinx

anonymous · Answer

sorry, i'm not sure where this is leading lol

rational · Answer

the question is actually tricky
how do you know the antiderivative of $\cos x$ is not $\sin x+100000$ ?

anonymous · Answer

Hmm.. well i guess it seems a little counter intuitive that the functions wouldn't be equal to each other given the same angle.  So then both answers are correct?

rational · Answer

Yes both answers are correct because they differ only by a constant.

rational · Answer

It would be fun to actually try and prove above statement

anonymous · Answer

Good ol' calculus.  Thanks for your help :)

rational · Answer

show that  "your answer" = "textbook answer" + C

anonymous · Answer

Lol it'll be fun when I don't have a final breathing down my neck.

rational · Answer

It is important to acknowledge that the antiderivative is not a single function, but a family of functions, all differing by a constant.

rational · Answer

$$\int \cos x\, dx = \sin x + C$$

where $C$ is an arbitrary constant

anonymous · Answer

Agreed - i like to know the why's.  Not just the how's.  Thanks for your time ! :)

sdfgsdfgs · Answer

@rational they are...cos^4 = (1-sin^2)^2
=1-2sin^2+sin^4
so -cos^4/4=1/4+sin^2/2-sin^4/4

rational · Answer

Ahh nice xD 
so not so hard to show they differ by a constant : 1/4

anonymous · Answer

This example illustrates a pitfall of following the textbook's solution.

campbell_st · Answer

well there seems to be a little problem in the interpretation of the question

i'd do this 

$$u = \cos(x) ~~~~\frac{du}{dx} = -\sin(x) ~~~~so  ~~~~du = -\sin(x)dx$$

so the problem, rewritten is 

$$\int\limits \cos^3(x) \sin(x)~dx$$

when when using substitution becomes

$$\int\limits u^3 \times -du!!~~~or ~~~-\int\limits u^3 du$$

which becomes $$-\frac{u^4}{4} + c$$

make the substitution for u and you get

$$-\frac{\cos^4(x)}{4}$$

this is a high school maths question and solution

campbell_st · Answer

hope it makes sense... 

it required substitution, but just a simple substitution.

campbell_st · Answer

oops forgot to add the constant C to the solution

anonymous · Answer

@campbell_st that would be the most direct substituion

anonymous · Answer

@campbell_st it makes perfect sense!  I'm beating myself up for not even considering that route...but this was overall a good exercise to see it worked out from different angles.