Expand integral dx/cosx
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Do some conjugate multiplication, apply some trig, and finish to get a result of ∫1cosx−1dx=cscx+cotx+C
Explanation:
As with most problems of this type, we'll solve it using a conjugate multiplication trick. Whenever you have something divided by something plus/minus something (as in 1cosx−1), it's always helpful to try conjugate multiplication, especially with trig functions.
We will begin by multiplying 1cosx−1 by the conjugate of cosx−1, which is cosx+1:
1cosx−1⋅cosx+1cosx+1
You may wonder why we do this. It's so we can apply the difference of squares property, (a−b)(a+b)=a2−b2, in the denominator, to simplify it a little. Back to the problem:
1cosx−1⋅cosx+1cosx+1=cosx+1(cosx−1)(cosx+1)
(cosx−1)(cosx+1))
IIIaXXXbXXXaXXXb
Notice how this is essentially (a−b)(a+b).
=cosx+1cos2x−1
Now, what about cos2x−1? Well, we know sin2x=1−cos2x. Let's multiply that by −1and see what we get:
−1(sin2x=1−cos2x)→−sin2x=−1+cos2x
=cos2−1
It turns out that −sin2x=cos2x−1, so let's replace cos2x−1:
cosx+1−sin2x
This is equivalent to cosx−sin2x+1−sin2x, which, using some trig, boils down to −cotxcscx−csc2x.
At this point, we've simplified to integral ∫1cosx−1dx to ∫−cotxcscx−csc2xdx. Using the sum rule, this becomes:
∫−cotxcscxdx+∫−csc2xdx
The first of these is cscx (because the derivative of cscx is −cotxcscx) and the second is cotx(because the derivative of cotx is −csc2x). Add on the constant of integration C and you have your solution:
∫1cosx−1dx=cscx+cotx+C
Explanation:
As with most problems of this type, we'll solve it using a conjugate multiplication trick. Whenever you have something divided by something plus/minus something (as in 1cosx−1), it's always helpful to try conjugate multiplication, especially with trig functions.
We will begin by multiplying 1cosx−1 by the conjugate of cosx−1, which is cosx+1:
1cosx−1⋅cosx+1cosx+1
You may wonder why we do this. It's so we can apply the difference of squares property, (a−b)(a+b)=a2−b2, in the denominator, to simplify it a little. Back to the problem:
1cosx−1⋅cosx+1cosx+1=cosx+1(cosx−1)(cosx+1)
(cosx−1)(cosx+1))
IIIaXXXbXXXaXXXb
Notice how this is essentially (a−b)(a+b).
=cosx+1cos2x−1
Now, what about cos2x−1? Well, we know sin2x=1−cos2x. Let's multiply that by −1and see what we get:
−1(sin2x=1−cos2x)→−sin2x=−1+cos2x
=cos2−1
It turns out that −sin2x=cos2x−1, so let's replace cos2x−1:
cosx+1−sin2x
This is equivalent to cosx−sin2x+1−sin2x, which, using some trig, boils down to −cotxcscx−csc2x.
At this point, we've simplified to integral ∫1cosx−1dx to ∫−cotxcscx−csc2xdx. Using the sum rule, this becomes:
∫−cotxcscxdx+∫−csc2xdx
The first of these is cscx (because the derivative of cscx is −cotxcscx) and the second is cotx(because the derivative of cotx is −csc2x). Add on the constant of integration C and you have your solution:
∫1cosx−1dx=cscx+cotx+C
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