Indefinite integral homework problem

[neshepard]

Homework Statement

∫cos(x)sin^6(x)*dx

Homework Equations

The Attempt at a Solution

u=sin(x) then du should be -cos(x) according to the integral tables?
-∫u^6*du = -u^7/7

-sin^7(x)/7+c yet my book and calculator tell me the answer is positive. How?

 

[Dick]

If u=sin(x), du=cos(x)*dx since (d/dx)sin(x)=+cos(x).

 

[neshepard]

That still doesn't make sense to me. According to our table of indefinite integrals,
∫sinx=-cosx + C. So how can d/dx make cosx positive? Especially when I continued in my homework and came across a problem with cosx/sin^2x and my answer which is correct by the book is -1/sinx where u=sinx and du=-cosx?

(Calc is too much of I don't like that 2+2=4 so I will play with the 2's till I get the answer to be 6)

Thanks

 

[Dick]

That still doesn't make sense to me. According to our table of indefinite integrals,
∫sinx=-cosx + C. So how can d/dx make cosx positive? Especially when I continued in my homework and came across a problem with cosx/sin^2x and my answer which is correct by the book is -1/sinx where u=sinx and du=-cosx?

(Calc is too much of I don't like that 2+2=4 so I will play with the 2's till I get the answer to be 6)

Thanks

Once you have chosen a u you differentiate it to find du. du=(du/dx)*dx. You don't integrate it. I.e if u=x^2, du=2xdx.

 

[neshepard]

So the handy dandy table is basically worthless unless I am to integrate exactly sinx?

Oh well. Thanks for the help.

 

[vela]

Especially when I continued in my homework and came across a problem with cosx/sin^2x and my answer which is correct by the book is -1/sinx where u=sinx and du=-cosx?

If you got the right answer, it's because you made another mistake which offset your first one.

(Calc is too much of I don't like that 2+2=4 so I will play with the 2's till I get the answer to be 6)

It's not uncommon for students learning calculus to feel this way because you're learning new techniques, but it is a sign that you don't understand what's going on completely. Rest assured everything fits together logically. There's no arbitrary application of ideas to change one answer into another.

So the handy dandy table is basically worthless unless I am to integrate exactly sin x?

The fundamental theorem of calculus tells you if

$$F(x) = \int f(x)\,dx$$

then

$$F'(x) = f(x)$$

So the integral of sin x you looked up in the table tells you (-cos x)' = sin x. It doesn't tell you anything about (sin x)', which is what you needed for this problem.

One handy mnemonic is that the derivatives of the "co" functions have the minus sign, so you have

$$\begin{align*}<br /> (\sin x)' & = \cos x \\<br /> (\tan x)' & = \sec^2 x \\<br /> (\sec x)' & = \sec x\tan x \\<br /> \\<br /> (\cos x)' & = -\sin x \\<br /> (\cot x)' & = -\csc^2 x \\<br /> (\csc x)' & = -\csc x\cot x<br /> \end{align*}$$

You're going to use these often, so it pays to just memorize them.