# Which two answers are correct?

• Dec 11th 2007, 04:53 AM
ebonyscythe
Which two answers are correct?
Here's the basic question:

We have an integral, that of which we do not know, where two out of three students got the correct answer. Which two students are right?

a. $sin^2x+C$
b. $-cos^2x+C$
c. $-sin^2x+C$

My logic is that one of the $(-)sin^2x+C$ is right, meaning the integral might be $\int2sin(x)cos(x)$, but I'm not sure how one could get a $-cos^2x+C$ out of that.

Any help is appreciated.
• Dec 11th 2007, 05:02 AM
Krizalid
An integral may have many primitives.

Why don't you differentiate the options to check if they yield the integrand.
• Dec 11th 2007, 05:13 AM
DivideBy0
Quote:

Originally Posted by Krizalid
An integral may have many primitives.

Why don't you differentiate the options to check if they yield the integrand.

When you talk about primitives, do you mean a given integral may have two answers which don't actually equal each other but are nonetheless valid evaluations of the integral?
• Dec 11th 2007, 05:26 AM
Soroban
Hello, ebonyscythe!

This is a classic problem . . .

Quote:

Integrate: . $2\int\sin x\cos x\,dx$
There are (at least) three solutions . . .

(1) Let $u = \sin x\quad\Rightarrow\quad du = \cos x\,dx$

Substitute: . $2\int u\,du \;=\;u^2 + C\;=\;\boxed{\sin^2\!x + C}$

(2) Let $u = \cos x\quad\Rightarrow\quad du = -\sin x\,dx$

Substitute: . $2\int u(-du) \:=\:-u^2 + C\:=\:\boxed{-\cos^2\!x + C}$

(3) . $\sin x\cos x \:=\:\frac{1}{2}(2\sin x\cos x) \;=\;\frac{1}{2}\sin2x$

So we have: . $2\int\frac{1}{2}\sin2x\,dx \;=\;\int\sin2x\,dx \;=\;\boxed{-\frac{1}{2}\cos2x + C}$

. . . . . and these three answers are equivalent.

• Dec 11th 2007, 05:48 AM
kalagota
if the choices were the answers of the 3 students and only two got the correct answer, then student who didn't get it right might had written $-\sin^2 x + C$, like what kriz wrote, differentiate the three, since we don't really know what the integral was, at least, when the three were differentiated, 2 of them would yield the same derivative..
• Dec 11th 2007, 05:49 AM
kalagota
Quote:

Originally Posted by DivideBy0
When you talk about primitives, do you mean a given integral may have two answers which don't actually equal each other but are nonetheless valid evaluations of the integral?

yes, they only differ by constants..
• Dec 11th 2007, 06:08 AM
Constatine11
Quote:

Originally Posted by ebonyscythe
Here's the basic question:

We have an integral, that of which we do not know, where two out of three students got the correct answer. Which two students are right?

a. $sin^2x+C$
b. $-cos^2x+C$
c. $-sin^2x+C$

My logic is that one of the $(-)sin^2x+C$ is right, meaning the integral might be $\int2sin(x)cos(x)$, but I'm not sure how one could get a $-cos^2x+C$ out of that.

Any help is appreciated.

Is it not the case that:

$\sin^2(x)=1-\cos^2(x)$?

ZB
• Dec 11th 2007, 06:33 AM
kalagota
Quote:

Originally Posted by Constatine11
Is it not the case that:

$\sin^2(x)=1-\cos^2(x)$?

ZB

it can be the case.. take note of choice B.. C takes charge the "1" in your right hand side..