Integral $intlimits_{sqrt{2}}^{2}frac1{t^3sqrt{t^2-1}}$
This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}}text{ d}ttext{.}$$
Because of the $sqrt{t^2-1} = sqrt{t^2-1^2}$ in the integrand, I set $t = 1sec(theta) = sec(theta)$, and $theta = sec^{-1}(t)$. Thus $text{d}t = sec(theta)tan(theta)text{ d}theta$ and
$$begin{align}intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}} text{ d}t&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{3}(theta)sqrt{sec^{2}(theta)-1}}sec(theta)tan(theta)text{ d}theta \
&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{2}(theta)}text{ d}theta \
&= intlimits_{pi/4}^{pi/3}cos^{2}(theta)text{ d}theta \
&= dfrac{1}{2}intlimits_{pi/4}^{pi/3}[1+cos(2theta)]text{ d}theta \
&= dfrac{1}{2}left{dfrac{pi}{3}-dfrac{pi}{4}+dfrac{1}{2}left[sinleft(dfrac{2pi}{3}right) - sinleft(dfrac{pi}{2}right)right]right}\
&= dfrac{1}{2}left[dfrac{pi}{12}+dfrac{1}{2}left(dfrac{sqrt{3}}{2}-1right)right] \
&= dfrac{pi}{24}+dfrac{1}{4}left(dfrac{sqrt{3}-2}{2}right)text{.}
end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $sin(x)$ memorized for $x = 0, dfrac{pi}{6}, dfrac{pi}{4}, dfrac{pi}{3}, dfrac{pi}{2}$, what is the easiest way to find $sinleft(dfrac{2pi}{3}right)$?
calculus integration proof-verification definite-integrals
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
add a comment |
This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}}text{ d}ttext{.}$$
Because of the $sqrt{t^2-1} = sqrt{t^2-1^2}$ in the integrand, I set $t = 1sec(theta) = sec(theta)$, and $theta = sec^{-1}(t)$. Thus $text{d}t = sec(theta)tan(theta)text{ d}theta$ and
$$begin{align}intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}} text{ d}t&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{3}(theta)sqrt{sec^{2}(theta)-1}}sec(theta)tan(theta)text{ d}theta \
&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{2}(theta)}text{ d}theta \
&= intlimits_{pi/4}^{pi/3}cos^{2}(theta)text{ d}theta \
&= dfrac{1}{2}intlimits_{pi/4}^{pi/3}[1+cos(2theta)]text{ d}theta \
&= dfrac{1}{2}left{dfrac{pi}{3}-dfrac{pi}{4}+dfrac{1}{2}left[sinleft(dfrac{2pi}{3}right) - sinleft(dfrac{pi}{2}right)right]right}\
&= dfrac{1}{2}left[dfrac{pi}{12}+dfrac{1}{2}left(dfrac{sqrt{3}}{2}-1right)right] \
&= dfrac{pi}{24}+dfrac{1}{4}left(dfrac{sqrt{3}-2}{2}right)text{.}
end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $sin(x)$ memorized for $x = 0, dfrac{pi}{6}, dfrac{pi}{4}, dfrac{pi}{3}, dfrac{pi}{2}$, what is the easiest way to find $sinleft(dfrac{2pi}{3}right)$?
calculus integration proof-verification definite-integrals
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
$sin(x) = cos(x - pi/2)$, then $sin(2pi/3) = cos(pi/6)$
– user67133
Oct 19 '14 at 1:38
@user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you!
– Clarinetist
Oct 19 '14 at 1:39
The (equivalent) identity $sin(x)=cos(pi/2-x)$ (and since $cos$ is even, we can get the previous one from there) might be easier to remember, since $pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement.
– Milo Brandt
Oct 19 '14 at 1:42
@Meelo - Clever! :O Thank you!
– Clarinetist
Oct 19 '14 at 1:43
add a comment |
This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}}text{ d}ttext{.}$$
Because of the $sqrt{t^2-1} = sqrt{t^2-1^2}$ in the integrand, I set $t = 1sec(theta) = sec(theta)$, and $theta = sec^{-1}(t)$. Thus $text{d}t = sec(theta)tan(theta)text{ d}theta$ and
$$begin{align}intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}} text{ d}t&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{3}(theta)sqrt{sec^{2}(theta)-1}}sec(theta)tan(theta)text{ d}theta \
&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{2}(theta)}text{ d}theta \
&= intlimits_{pi/4}^{pi/3}cos^{2}(theta)text{ d}theta \
&= dfrac{1}{2}intlimits_{pi/4}^{pi/3}[1+cos(2theta)]text{ d}theta \
&= dfrac{1}{2}left{dfrac{pi}{3}-dfrac{pi}{4}+dfrac{1}{2}left[sinleft(dfrac{2pi}{3}right) - sinleft(dfrac{pi}{2}right)right]right}\
&= dfrac{1}{2}left[dfrac{pi}{12}+dfrac{1}{2}left(dfrac{sqrt{3}}{2}-1right)right] \
&= dfrac{pi}{24}+dfrac{1}{4}left(dfrac{sqrt{3}-2}{2}right)text{.}
end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $sin(x)$ memorized for $x = 0, dfrac{pi}{6}, dfrac{pi}{4}, dfrac{pi}{3}, dfrac{pi}{2}$, what is the easiest way to find $sinleft(dfrac{2pi}{3}right)$?
calculus integration proof-verification definite-integrals
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}}text{ d}ttext{.}$$
Because of the $sqrt{t^2-1} = sqrt{t^2-1^2}$ in the integrand, I set $t = 1sec(theta) = sec(theta)$, and $theta = sec^{-1}(t)$. Thus $text{d}t = sec(theta)tan(theta)text{ d}theta$ and
$$begin{align}intlimits_{sqrt{2}}^{2}dfrac{1}{t^3sqrt{t^2-1}} text{ d}t&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{3}(theta)sqrt{sec^{2}(theta)-1}}sec(theta)tan(theta)text{ d}theta \
&= intlimits_{pi/4}^{pi/3}dfrac{1}{sec^{2}(theta)}text{ d}theta \
&= intlimits_{pi/4}^{pi/3}cos^{2}(theta)text{ d}theta \
&= dfrac{1}{2}intlimits_{pi/4}^{pi/3}[1+cos(2theta)]text{ d}theta \
&= dfrac{1}{2}left{dfrac{pi}{3}-dfrac{pi}{4}+dfrac{1}{2}left[sinleft(dfrac{2pi}{3}right) - sinleft(dfrac{pi}{2}right)right]right}\
&= dfrac{1}{2}left[dfrac{pi}{12}+dfrac{1}{2}left(dfrac{sqrt{3}}{2}-1right)right] \
&= dfrac{pi}{24}+dfrac{1}{4}left(dfrac{sqrt{3}-2}{2}right)text{.}
end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $sin(x)$ memorized for $x = 0, dfrac{pi}{6}, dfrac{pi}{4}, dfrac{pi}{3}, dfrac{pi}{2}$, what is the easiest way to find $sinleft(dfrac{2pi}{3}right)$?
calculus integration proof-verification definite-integrals
calculus integration proof-verification definite-integrals
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:16
Martin Sleziak
44.7k7115270
44.7k7115270
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
asked Oct 19 '14 at 1:35
Clarinetist
10.8k42778
10.8k42778
$sin(x) = cos(x - pi/2)$, then $sin(2pi/3) = cos(pi/6)$
– user67133
Oct 19 '14 at 1:38
@user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you!
– Clarinetist
Oct 19 '14 at 1:39
The (equivalent) identity $sin(x)=cos(pi/2-x)$ (and since $cos$ is even, we can get the previous one from there) might be easier to remember, since $pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement.
– Milo Brandt
Oct 19 '14 at 1:42
@Meelo - Clever! :O Thank you!
– Clarinetist
Oct 19 '14 at 1:43
add a comment |
$sin(x) = cos(x - pi/2)$, then $sin(2pi/3) = cos(pi/6)$
– user67133
Oct 19 '14 at 1:38
@user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you!
– Clarinetist
Oct 19 '14 at 1:39
The (equivalent) identity $sin(x)=cos(pi/2-x)$ (and since $cos$ is even, we can get the previous one from there) might be easier to remember, since $pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement.
– Milo Brandt
Oct 19 '14 at 1:42
@Meelo - Clever! :O Thank you!
– Clarinetist
Oct 19 '14 at 1:43
$sin(x) = cos(x - pi/2)$, then $sin(2pi/3) = cos(pi/6)$
– user67133
Oct 19 '14 at 1:38
$sin(x) = cos(x - pi/2)$, then $sin(2pi/3) = cos(pi/6)$
– user67133
Oct 19 '14 at 1:38
@user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you!
– Clarinetist
Oct 19 '14 at 1:39
@user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you!
– Clarinetist
Oct 19 '14 at 1:39
The (equivalent) identity $sin(x)=cos(pi/2-x)$ (and since $cos$ is even, we can get the previous one from there) might be easier to remember, since $pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement.
– Milo Brandt
Oct 19 '14 at 1:42
The (equivalent) identity $sin(x)=cos(pi/2-x)$ (and since $cos$ is even, we can get the previous one from there) might be easier to remember, since $pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement.
– Milo Brandt
Oct 19 '14 at 1:42
@Meelo - Clever! :O Thank you!
– Clarinetist
Oct 19 '14 at 1:43
@Meelo - Clever! :O Thank you!
– Clarinetist
Oct 19 '14 at 1:43
add a comment |
2 Answers
2
active
oldest
votes
Your working looks alright.$$sin theta= sin (pi - theta)$$
Is another identity.
edited Dec 3 at 2:51
Frank W.
2,9561316
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
add a comment |
There are no issues with your solution.
You can easily find $sin frac{2pi}{3}$ using $sin x = sin (pi - x)$ for $x = frac{pi}{3}$.
answered Oct 19 '14 at 1:40
James Harrison
834615
add a comment |
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2 Answers
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2 Answers
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Your working looks alright.$$sin theta= sin (pi - theta)$$
Is another identity.
edited Dec 3 at 2:51
Frank W.
2,9561316
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
add a comment |
Your working looks alright.$$sin theta= sin (pi - theta)$$
Is another identity.
edited Dec 3 at 2:51
Frank W.
2,9561316
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
add a comment |
Your working looks alright.$$sin theta= sin (pi - theta)$$
Is another identity.
edited Dec 3 at 2:51
Frank W.
2,9561316
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
Your working looks alright.$$sin theta= sin (pi - theta)$$
Is another identity.
edited Dec 3 at 2:51
Frank W.
2,9561316
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
edited Dec 3 at 2:51
Frank W.
2,9561316
edited Dec 3 at 2:51
Frank W.
2,9561316
edited Dec 3 at 2:51
Frank W.
2,9561316
2,9561316
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
answered Oct 19 '14 at 1:40
Sherlock Holmes
1,788713
1,788713
add a comment |
add a comment |
There are no issues with your solution.
You can easily find $sin frac{2pi}{3}$ using $sin x = sin (pi - x)$ for $x = frac{pi}{3}$.
answered Oct 19 '14 at 1:40
James Harrison
834615
add a comment |
There are no issues with your solution.
You can easily find $sin frac{2pi}{3}$ using $sin x = sin (pi - x)$ for $x = frac{pi}{3}$.
answered Oct 19 '14 at 1:40
James Harrison
834615
add a comment |
There are no issues with your solution.
You can easily find $sin frac{2pi}{3}$ using $sin x = sin (pi - x)$ for $x = frac{pi}{3}$.
answered Oct 19 '14 at 1:40
James Harrison
834615
There are no issues with your solution.
You can easily find $sin frac{2pi}{3}$ using $sin x = sin (pi - x)$ for $x = frac{pi}{3}$.
answered Oct 19 '14 at 1:40
James Harrison
834615
answered Oct 19 '14 at 1:40
James Harrison
834615
answered Oct 19 '14 at 1:40
James Harrison
834615
answered Oct 19 '14 at 1:40
James Harrison
834615
834615
add a comment |
add a comment |
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$sin(x) = cos(x - pi/2)$, then $sin(2pi/3) = cos(pi/6)$
– user67133
Oct 19 '14 at 1:38
@user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you!
– Clarinetist
Oct 19 '14 at 1:39
The (equivalent) identity $sin(x)=cos(pi/2-x)$ (and since $cos$ is even, we can get the previous one from there) might be easier to remember, since $pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement.
– Milo Brandt
Oct 19 '14 at 1:42
@Meelo - Clever! :O Thank you!
– Clarinetist
Oct 19 '14 at 1:43