Integrate sin x / (cos x)^3 ?

Question

Thanks Pascal and C-Wryte. I used both approaches since the derivative of tan x is sec(x)^2 and the derivative of sec x is tan x sec x. But I can't see the equivalence of 1/2tan(x)^2 and 1/2 sec(x)^2?

Answer 1

∫sin x/cos³ x dx

Rewrite this in terms of tan and sec:

∫tan x sec² x dx

Now using the inverse chain rule:

1/2 tan² x + C

And we are done.

Edited to add: The answerers below me have helped demonstrate an important point -- that C is important! Depending on which method you use, it is possible to obtain two equally simple antiderivatives of a function which are not the same. However, assuming you have done your work correctly, the answers will always differ by a constant (in this case, 1/2 tan² x = 1/2 sec² x + 1/2). The C then, accounts for the difference between the possible antiderivatives of a function. If you didn't account for that constant, you might accidentally reason that 1/2 tan² x = ∫sin x/cos³ x dx = 1/2 sec² x → 1/2 = 0, and thereby cause confusion.

Further addition: the equivalence is not between 1/2 tan² x and 1/2 sec² x, it is between 1/2 tan² x + C and 1/2 sec² x + C. That C is important. Any function you can obtain by adding some constant to 1/2 sec² x can be obtained by adding some (different) constant to 1/2 tan² x. So to obtain 1/2 sec² x from 1/2 tan² x + C, we would let C=1/2, since 1/2 tan² x + 1/2 = (tan² x + 1)/2 = (sec² x)/2. Conversely, to obtain 1/2 tan² x from 1/2 sec² x + C, we would let C=-1/2. This follows from the trigonometric identity tan² x + 1 = sec² x (if you don't know this identity, it follows since tan² x + 1 = sin² x/cos² x + 1 = sin² x/cos² x + cos² x/cos² x = (sin² x + cos² x)/cos² x = 1/cos² x = sec² x).

If it still doesn't make any sense, consider that 1/2 tan² x + C doesn't refer to a single function, but rather an entire set of functions, one for each value of C. Similarly, 1/2 sec² x + C refers to an entire set of functions, one for each value of C. And since these may be different constants in each case, we can label them differently, say, C₁ and C₂. So for each value of C₁, there is some value of C₂ I can choose (specifically, C₂ = C₁-1/2) such that 1/2 tan² x + C₁ = 1/2 sec² + C₂, and vice versa. That is why the two solutions are equivalent.

Answer 2

∫sin x/cos³ x dx

You must first note that this could also be written in terms of tangent and secant, or

∫tan x sec² x dx = ∫tan x sec x sec x dx

Now, we know that the derivative of sec x is sec x tan x, so we can define:

u = sec x
du = sec x tan x

Which simplifies the integral to:

∫u du = ½u² + C

So, plug back in sec x for the u, and you have your final answer of

½ sec² x + C
__________

Answer 3

I agree with the 1st post in that you need to use inverse chain rule, by I think he made a slip in one of his missing steps.

1st thing you should notice is that sin(x) is a constant multiple of the derivative of cos x. This means that it might be fun to call u = cos(x) and du/dx = -sin(x).

So we have integral{ sin(x) dx / (cos(x))^3 } = integral { - du / u^3 }

So after a simple use of the power rule...
integral{ sin(x) dx / (cos(x))^3 } = 1 / (2 u^2) = 1/2 (sec x)^2

Which you can verify by taking the derivative of this easy answer.

Answer 4

I = ∫ sin x / [ cos x.(cos ² x) ]
I = ∫ tan x . sec ² x dx
let u = tan x
du = sec ² x dx
I = ∫ u du
I = u ² / 2 + C
I = tan ² x / 2 + C

Answer 5

s:integral sign
-s -sin(cosx)^-3= (-cos^-2)/-2=(cos^-2)/2+c
1/2cos^2+c