# u-substitution, definite integral, trig function

#### NecroWinter

##### New member
problem: sin(x)^2cos(x)
integrate from 0 to pi/2

the way I tried this is by making u = sin, and then working with du to make cos cancel out
the next thing I did was check my new intervals by plugging in pi/2 and 0 into the u function (becomes 0 to 1)
next, I took the anti derivative of u, which came to (u^3)/3

((u^3)/3))-((u^3)/3)
sin0=0
sin pi/2 = 1

(0^3)/3 - (1^3)/3 = -1/3

except, wolfram alpha gives me 1/3.

What happened?

#### Krizalid

##### Active member
Re: usubstitution, definite integral, trig function

You actually evaluate first 1 and later 0.

Cheap observation: we're integrating a non-negative function. How could have a negative integral?

#### Prove It

##### Well-known member
MHB Math Helper
Re: usubstitution, definite integral, trig function

problem: sin(x)^2cos(x)
integrate from 0 to pi/2

the way I tried this is by making u = sin, and then working with du to make cos cancel out
the next thing I did was check my new intervals by plugging in pi/2 and 0 into the u function (becomes 0 to 1)
next, I took the anti derivative of u, which came to (u^3)/3

((u^3)/3))-((u^3)/3)
sin0=0
sin pi/2 = 1

(0^3)/3 - (1^3)/3 = -1/3

except, wolfram alpha gives me 1/3.

What happened?
OK, first of all, $\displaystyle \int{u^2\,du}$ is NOT $\displaystyle \frac{u^3}{3} - \frac{u^3}{3}$, it's just $\displaystyle \frac{u^3}{3} + C$.

Second, Krizalid is correct in stating that the definite integral is evaluated by subtracting the lower limit from the upper limit, not the other way around as you have done.

#### NecroWinter

##### New member
Re: usubstitution, definite integral, trig function

OK, first of all, $\displaystyle \int{u^2\,du}$ is NOT $\displaystyle \frac{u^3}{3} - \frac{u^3}{3}$, it's just $\displaystyle \frac{u^3}{3} + C$.

Second, Krizalid is correct in stating that the definite integral is evaluated by subtracting the lower limit from the upper limit, not the other way around as you have done.
im aware that the antiderivative is that, but what you are citing is me skipping a step and applying the fundamental theorem of calculus before entering the numbers.

#### Prove It

##### Well-known member
MHB Math Helper
Re: usubstitution, definite integral, trig function

im aware that the antiderivative is that, but what you are citing is me skipping a step and applying the fundamental theorem of calculus before entering the numbers.
Skipping steps is one thing, writing something that is incorrect is another thing entirely.

What you should have written is this...

\displaystyle \begin{align*} \int_0^1{u^2\,du} &= \left[\frac{u^3}{3}\right]_0^1 \\ &= \frac{1^3}{3} - \frac{0^3}{3} \end{align*}

#### CaptainBlack

##### Well-known member
Re: usubstitution, definite integral, trig function

problem: sin(x)^2cos(x)
integrate from 0 to pi/2

the way I tried this is by making u = sin, and then working with du to make cos cancel out
the next thing I did was check my new intervals by plugging in pi/2 and 0 into the u function (becomes 0 to 1)
next, I took the anti derivative of u, which came to (u^3)/3

((u^3)/3))-((u^3)/3)
sin0=0
sin pi/2 = 1

(0^3)/3 - (1^3)/3 = -1/3

except, wolfram alpha gives me 1/3.

What happened?
Unrelated to what everyone else has posted, but you should always look at the integrand to see if it is an obvious derivative. In this case the integrand is obviously the derivative of $$\frac{1}{3}\sin^3(x)$$.

CB

#### tomatballooncal

##### New member
Re: usubstitution, definite integral, trig function

Just in case a picture helps... ... where (key in spoiler) ... ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is... ________________________________________________________

Don't integrate - balloontegrate!

Hi guys - most grateful for your toleration if it still extends... any chance of enabling links? Or a way to enlarge pics? And is there a spoiler code?

#### Attachments

• 14.3 KB Views: 6
• 16.4 KB Views: 7
• CaptainBlack

#### NecroWinter

##### New member
Re: usubstitution, definite integral, trig function

Skipping steps is one thing, writing something that is incorrect is another thing entirely.

What you should have written is this...

\displaystyle \begin{align*} \int_0^1{u^2\,du} &= \left[\frac{u^3}{3}\right]_0^1 \\ &= \frac{1^3}{3} - \frac{0^3}{3} \end{align*}
I wrote:
"I took the anti derivative of u, which came to (u^3)/3"

the part you are quoting is being misunderstood.

thanks everyone

#### HallsofIvy

##### Well-known member
MHB Math Helper
Re: usubstitution, definite integral, trig function

I wrote:
"I took the anti derivative of u, which came to (u^3)/3"

the part you are quoting is being misunderstood.

thanks everyone
Yes, and the part that was quoted was $\frac{u^3}{3}- \frac{u^3}{3}$.
I hope you understand that that is very bad notation that is easily misunderstood.
What did you mean by that?
In any case, if F is an anti-derivative of f, then
$\int_a^b f(x)dx= F(b)- F(a)$, not F(a)- F(b).