Absolute value in indefinite integral
I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$
I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$
I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?
Thanks in advance.
indefinite-integrals absolute-value
edited Dec 21 at 17:40
trancelocation
9,0851521
asked Dec 21 at 8:53
user121
273
add a comment |
I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$
I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$
I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?
Thanks in advance.
indefinite-integrals absolute-value
edited Dec 21 at 17:40
trancelocation
9,0851521
asked Dec 21 at 8:53
user121
273
1
Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58
2
Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22
add a comment |
I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$
I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$
I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?
Thanks in advance.
indefinite-integrals absolute-value
edited Dec 21 at 17:40
trancelocation
9,0851521
asked Dec 21 at 8:53
user121
273
I have to show that $$ int left(frac{dx}{x^2 sqrt{x^2+4}}right) = left(frac{-sqrt{x^2+4}}{4x}right) + c$$
I used the substitution $ frac{x}{2} = tan u$, and I got:
$$frac{1}{4}int left(frac{ |cos u|; du}{ (sin u)^2 }right)$$
I saw in the solution of this task that $|cos u| = cos u$. Why do we ignore the absolute value?
Thanks in advance.
indefinite-integrals absolute-value
indefinite-integrals absolute-value
edited Dec 21 at 17:40
trancelocation
9,0851521
asked Dec 21 at 8:53
user121
273
edited Dec 21 at 17:40
trancelocation
9,0851521
asked Dec 21 at 8:53
user121
273
edited Dec 21 at 17:40
trancelocation
9,0851521
edited Dec 21 at 17:40
trancelocation
9,0851521
edited Dec 21 at 17:40
trancelocation
9,0851521
9,0851521
asked Dec 21 at 8:53
user121
273
asked Dec 21 at 8:53
user121
273
asked Dec 21 at 8:53
user121
273
273
1
Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58
2
Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22
add a comment |
1
Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58
2
Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22
1
1
Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58
Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58
2
2
Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22
Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22
add a comment |
2 Answers
2
active
oldest
votes
When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
add a comment |
Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.
Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?
answered Dec 21 at 9:00
KM101
4,418418
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
add a comment |
When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
add a comment |
When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
When defining $x=tan u$ and from $$-infty<x=tan u<infty$$we conclude that $$-{pi over 2}<u<{pi over 2}$$is sufficient. In this interval $0<cos ule1$ and $$|cos u|=cos u$$
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
answered Dec 21 at 11:18
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.
Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?
answered Dec 21 at 9:00
KM101
4,418418
add a comment |
Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.
Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?
answered Dec 21 at 9:00
KM101
4,418418
add a comment |
Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.
Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?
answered Dec 21 at 9:00
KM101
4,418418
Note that the domain of $tan (u)$ is $u in big(-frac{pi}{2}, frac{pi}{2}big)$.
Where is $cos(u)$ positive? What can you conclude from that regarding $vert cos(u)vert$ and $cos(u)$?
answered Dec 21 at 9:00
KM101
4,418418
answered Dec 21 at 9:00
KM101
4,418418
answered Dec 21 at 9:00
KM101
4,418418
answered Dec 21 at 9:00
KM101
4,418418
4,418418
add a comment |
add a comment |
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1
Cause u is between $-pi/2, pi/2$
– dmtri
Dec 21 at 8:58
2
Being lazy, I computed the derivative of the rhs, simplified and got the integrand.
– Claude Leibovici
Dec 21 at 9:22