Sine from negative cosine
$begingroup$
I'm trying to solve the integral $I=int frac{dx}{x^3 sqrt{x^2-4}}$ ($x<0$) which can be solved by substituting $x=2 sec t$. One ends up with $$I=-frac{1}{16} left( t+sin t cos t right)+C$$
(This is verified by the book I'm using.)
If $x=2 sec t$, then $t=sec^{-1}(frac x2)$ and $cos t =frac 2x$.
My problem is replacing $sin t$ by something which is dependent on $x$. The standard method would be constructing a right-angled triangle with sides given by $cos t=frac 2x$, giving $sin t$. However, since $x<0$, the triangle would have negative sides which is causing me trouble.
The solution given in the book is $sin t = frac{-sqrt{x^2-4}}{x}$ but I don't see what justifies the sign. (Could someone explain it?) Another method would be using the identity $sin^2 t+cos^2 t = 1$ but this also only gives $sin t = pm sqrt{1-frac{4} {x^2}}$.
integration substitution
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
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add a comment |
$begingroup$
I'm trying to solve the integral $I=int frac{dx}{x^3 sqrt{x^2-4}}$ ($x<0$) which can be solved by substituting $x=2 sec t$. One ends up with $$I=-frac{1}{16} left( t+sin t cos t right)+C$$
(This is verified by the book I'm using.)
If $x=2 sec t$, then $t=sec^{-1}(frac x2)$ and $cos t =frac 2x$.
My problem is replacing $sin t$ by something which is dependent on $x$. The standard method would be constructing a right-angled triangle with sides given by $cos t=frac 2x$, giving $sin t$. However, since $x<0$, the triangle would have negative sides which is causing me trouble.
The solution given in the book is $sin t = frac{-sqrt{x^2-4}}{x}$ but I don't see what justifies the sign. (Could someone explain it?) Another method would be using the identity $sin^2 t+cos^2 t = 1$ but this also only gives $sin t = pm sqrt{1-frac{4} {x^2}}$.
integration substitution
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the integral $I=int frac{dx}{x^3 sqrt{x^2-4}}$ ($x<0$) which can be solved by substituting $x=2 sec t$. One ends up with $$I=-frac{1}{16} left( t+sin t cos t right)+C$$
(This is verified by the book I'm using.)
If $x=2 sec t$, then $t=sec^{-1}(frac x2)$ and $cos t =frac 2x$.
My problem is replacing $sin t$ by something which is dependent on $x$. The standard method would be constructing a right-angled triangle with sides given by $cos t=frac 2x$, giving $sin t$. However, since $x<0$, the triangle would have negative sides which is causing me trouble.
The solution given in the book is $sin t = frac{-sqrt{x^2-4}}{x}$ but I don't see what justifies the sign. (Could someone explain it?) Another method would be using the identity $sin^2 t+cos^2 t = 1$ but this also only gives $sin t = pm sqrt{1-frac{4} {x^2}}$.
integration substitution
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
$endgroup$
I'm trying to solve the integral $I=int frac{dx}{x^3 sqrt{x^2-4}}$ ($x<0$) which can be solved by substituting $x=2 sec t$. One ends up with $$I=-frac{1}{16} left( t+sin t cos t right)+C$$
(This is verified by the book I'm using.)
If $x=2 sec t$, then $t=sec^{-1}(frac x2)$ and $cos t =frac 2x$.
My problem is replacing $sin t$ by something which is dependent on $x$. The standard method would be constructing a right-angled triangle with sides given by $cos t=frac 2x$, giving $sin t$. However, since $x<0$, the triangle would have negative sides which is causing me trouble.
The solution given in the book is $sin t = frac{-sqrt{x^2-4}}{x}$ but I don't see what justifies the sign. (Could someone explain it?) Another method would be using the identity $sin^2 t+cos^2 t = 1$ but this also only gives $sin t = pm sqrt{1-frac{4} {x^2}}$.
integration substitution
integration substitution
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
edited Jan 7 at 13:13
FizzleDizzle
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
asked Jan 7 at 11:55
FizzleDizzleFizzleDizzle
895
895
add a comment |
add a comment |
1 Answer
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If $-frac{1}{2}pi leq tleq 0$, then $-1 leq sin t leq 0$, and all values in this range are achieved. Therefore, since $-1 < -frac{sqrt{x^2-4}}x leq 0$, there exist valid values of $t$ making this formula true, so you are allowed to make the substitution, and we could choose $t$ with $-frac12pileq tleq 0$. There doesn't need to be an actual triangle, the function does take negative values for these arguments.
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
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1 Answer
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1 Answer
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$begingroup$
If $-frac{1}{2}pi leq tleq 0$, then $-1 leq sin t leq 0$, and all values in this range are achieved. Therefore, since $-1 < -frac{sqrt{x^2-4}}x leq 0$, there exist valid values of $t$ making this formula true, so you are allowed to make the substitution, and we could choose $t$ with $-frac12pileq tleq 0$. There doesn't need to be an actual triangle, the function does take negative values for these arguments.
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
$endgroup$
add a comment |
$begingroup$
If $-frac{1}{2}pi leq tleq 0$, then $-1 leq sin t leq 0$, and all values in this range are achieved. Therefore, since $-1 < -frac{sqrt{x^2-4}}x leq 0$, there exist valid values of $t$ making this formula true, so you are allowed to make the substitution, and we could choose $t$ with $-frac12pileq tleq 0$. There doesn't need to be an actual triangle, the function does take negative values for these arguments.
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
$endgroup$
add a comment |
$begingroup$
If $-frac{1}{2}pi leq tleq 0$, then $-1 leq sin t leq 0$, and all values in this range are achieved. Therefore, since $-1 < -frac{sqrt{x^2-4}}x leq 0$, there exist valid values of $t$ making this formula true, so you are allowed to make the substitution, and we could choose $t$ with $-frac12pileq tleq 0$. There doesn't need to be an actual triangle, the function does take negative values for these arguments.
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
$endgroup$
If $-frac{1}{2}pi leq tleq 0$, then $-1 leq sin t leq 0$, and all values in this range are achieved. Therefore, since $-1 < -frac{sqrt{x^2-4}}x leq 0$, there exist valid values of $t$ making this formula true, so you are allowed to make the substitution, and we could choose $t$ with $-frac12pileq tleq 0$. There doesn't need to be an actual triangle, the function does take negative values for these arguments.
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
edited Jan 7 at 12:33
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
answered Jan 7 at 12:22
Matt SamuelMatt Samuel
37.5k63665
37.5k63665
add a comment |
add a comment |
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