Which two answers are correct?

**ebonyscythe** · December 11th 2007, 04:53 AM

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.

**Krizalid** · December 11th 2007, 05:02 AM

An integral may have many primitives.

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

**DivideBy0** · December 11th 2007, 05:13 AM

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?

**Soroban** · December 11th 2007, 05:26 AM

Hello, ebonyscythe!

This is a classic problem . . .

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.

**kalagota** · December 11th 2007, 05:48 AM

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..

**kalagota** · December 11th 2007, 05:49 AM

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..

**Constatine11** · December 11th 2007, 06:08 AM

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

**kalagota** · December 11th 2007, 06:33 AM

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..

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