integrate sin x/cos^3 x dx?

Question

What integral rules do I need to use to solve those kind of probems? ( integrate sin x/cos^3 x dx )

Answer 1

There are different ways to solve this, but here's the way I'd go about it:

∫ sin(x) / cos^3 (x) dx =
∫ sin(x) [ cos (x) ]^-3 dx =
(-1/2) ∫ sin(x) * -2[ cos (x) ]^-3 dx =
(1/2) ∫ -sin(x) * -2[ cos (x) ]^-3 dx =
(1/2) cos^-2 (x) + c

If you use "u substitition" by letting u = cos(x) so that du = -sin(x) dx, etc. you'll get the right answer. I just noticed that you have cos(x) raised to some power, and that the derivative of cosine is -sin(x), so I did the "chain rule" in reverse.

Somebody else answered (1/2)tan²(x) + c, which is not wrong. If you use the fact that sin^2 (x) + cos^2(x) = 1, and apply it to the other answer, then you have:
(1/2) (1/cos^2 (x)) + c =
(1/2) [sin^2 (x) + cos^2(x)]/cos^2 (x)) + c =
(1/2) [tan^2 (x) + 1] + c =
(1/2)tan^2 (x) + 1/2 + c =
(1/2)tan^2 (x) + (1/2 + c) =
(1/2)tan^2 (x) + (a constant)

Answer 2

Since the derivative of tan(x) is sec²(x),
∫tan(x) sec²(x) dx
= (1/2)tan²(x) + C

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Update: the reason that you are getting so many different looking answers below is because 1+tan²(x) = sec²(x). In particular, suppose that C = 1/2 + D
Then the answer would be
(1/2)tan²(x) + 1/2 + D
= (1/2)(tan²(x) + 1) + D
= (1/2)sec²(x) + D
= (1/2)(1/cos²(x)) + D
These answers are all equivalent.

It's one of the few cases where that constant you always tack on after integration affects what the result looks like. So in general, to check if two integrations of the same function are equivalent, you should subtract the two results and if the difference is a constant, you know that both answers are just as valid (or just as wrong, for that matter).

Answer 3

I think all of the answers below miss the point.

This particular integral is one of a very powerful group which is often overlooked by students.

It belongs to a surprisingly common group of the type

Int [ f '()x)/ f(ish)(x) ] dx. I wish I could explain ish

You have Int sin cos^(-3)x dx and so the answer will be of the form cos^(-2)x because its derivative is 2(cos(^(-3)) x. sinx

The answer to your integral will be cos^(-2)x / 2 + c

Answer 4

Let cos x=u
-sinxdx=du
sinxdx=-du

integral -u^(-3) du
=1/(2*u^2)

=1/(2*cos^2 x)+C

Answer 5

1) Go to google.

2) Type in 'integrator'.

3) Click 'I'm Feeling Lucky'.

Works every time.

Answer 6

The first guy's answer is
(1/2) tan^2 x + constant

The second guy's answer is
(1/2) (1/cos^2 x) + constant

Yet they are both correct. Can you tell me why?

Answer 7

You need to do a u-substitution. Let u = cos x, then du = -sinx dx OR -du = sinx dx

So you really have

-∫u^-3 du

= 0.5 u^-2 + c
= 0.5 (cos x)^-2 + c
= .5 sec²x + c

Answer 8

I = ∫sinx / (cos³x) dx
Let u = cos x
du = - sin x.dx
- du = sin x.dx
I = - ∫ du / u³
I = - ∫ u^(-3).du
I = (1/2u²) + C
I = 1 / 2 cos ² x + C
I = (1/2).sec ² x + C

Answer 9

Err wha?

Answer 10

the point of ur maths homework is that u do it urself, y'know