Integrate using trigonometric substitution. Am I on the right path?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I have been trying to solve:
$$int fracsqrtx^2-9x^3 dx$$
I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$
So then I have:
$$int fracsqrt9sec^2 theta - 927 sec^3 theta dx$$
$$int fracsqrt9(sec^2 theta - 1)27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ int frac3tan theta27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ = int frac9 tan ^2 theta27 sec ^2 theta d theta$$
$$ = int fractan ^2 theta3 sec ^2 theta d theta$$
$$ int fracsin^2 thetacos^2 theta cdot fraccos^2 theta3 d theta$$
$$ int fracsin^2 theta3 d theta$$
$$frac13 int sin^2 theta d theta$$
Am I on the right track?
integration
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
$endgroup$
add a comment |
$begingroup$
I have been trying to solve:
$$int fracsqrtx^2-9x^3 dx$$
I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$
So then I have:
$$int fracsqrt9sec^2 theta - 927 sec^3 theta dx$$
$$int fracsqrt9(sec^2 theta - 1)27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ int frac3tan theta27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ = int frac9 tan ^2 theta27 sec ^2 theta d theta$$
$$ = int fractan ^2 theta3 sec ^2 theta d theta$$
$$ int fracsin^2 thetacos^2 theta cdot fraccos^2 theta3 d theta$$
$$ int fracsin^2 theta3 d theta$$
$$frac13 int sin^2 theta d theta$$
Am I on the right track?
integration
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
$endgroup$
$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46
1
$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06
add a comment |
$begingroup$
I have been trying to solve:
$$int fracsqrtx^2-9x^3 dx$$
I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$
So then I have:
$$int fracsqrt9sec^2 theta - 927 sec^3 theta dx$$
$$int fracsqrt9(sec^2 theta - 1)27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ int frac3tan theta27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ = int frac9 tan ^2 theta27 sec ^2 theta d theta$$
$$ = int fractan ^2 theta3 sec ^2 theta d theta$$
$$ int fracsin^2 thetacos^2 theta cdot fraccos^2 theta3 d theta$$
$$ int fracsin^2 theta3 d theta$$
$$frac13 int sin^2 theta d theta$$
Am I on the right track?
integration
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
$endgroup$
I have been trying to solve:
$$int fracsqrtx^2-9x^3 dx$$
I am letting $ x = 3sec theta$ and so $dx = 3 sec theta tan theta$
So then I have:
$$int fracsqrt9sec^2 theta - 927 sec^3 theta dx$$
$$int fracsqrt9(sec^2 theta - 1)27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ int frac3tan theta27 sec^3 theta 3 sec theta tan theta, d theta$$
$$ = int frac9 tan ^2 theta27 sec ^2 theta d theta$$
$$ = int fractan ^2 theta3 sec ^2 theta d theta$$
$$ int fracsin^2 thetacos^2 theta cdot fraccos^2 theta3 d theta$$
$$ int fracsin^2 theta3 d theta$$
$$frac13 int sin^2 theta d theta$$
Am I on the right track?
integration
integration
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
edited Feb 3 at 17:17
J. W. Tanner
2,4831117
2,4831117
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
asked Feb 3 at 16:36
Jwan622Jwan622
2,18911631
2,18911631
$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46
1
$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06
add a comment |
$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46
1
$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06
$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46
$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46
1
1
$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06
$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Yes, your solution so far is correct. Now, to integrate $int sin^2theta,dtheta$, use the half-angle formula for the sine function:
$$
sin^2theta=frac1-cos(2theta)2.
$$
Also, a bit later, you're going to need this formula:
$$
sin(2theta)=2sinthetacostheta.
$$
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
$endgroup$
add a comment |
$begingroup$
$$I=intfracsqrtx^2-9x^3dx=intfrac3sqrt(x/3)^2-1x^3dx$$
and we know that $cosh^2theta-1=sinh^2theta$
$$x=3cosh(y)$$
$$dx=3sinh(y)dy$$
$$I=9intfracsqrtcosh^2(y)-127cosh^3(y)sinh(y)dy=frac13intfracsinh^2(y)cosh^3(y)dy=frac13inttextsech(y)-textsech^3(y)dy$$
a reduction formula can then be used
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, your solution so far is correct. Now, to integrate $int sin^2theta,dtheta$, use the half-angle formula for the sine function:
$$
sin^2theta=frac1-cos(2theta)2.
$$
Also, a bit later, you're going to need this formula:
$$
sin(2theta)=2sinthetacostheta.
$$
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
$endgroup$
add a comment |
$begingroup$
Yes, your solution so far is correct. Now, to integrate $int sin^2theta,dtheta$, use the half-angle formula for the sine function:
$$
sin^2theta=frac1-cos(2theta)2.
$$
Also, a bit later, you're going to need this formula:
$$
sin(2theta)=2sinthetacostheta.
$$
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
$endgroup$
add a comment |
$begingroup$
Yes, your solution so far is correct. Now, to integrate $int sin^2theta,dtheta$, use the half-angle formula for the sine function:
$$
sin^2theta=frac1-cos(2theta)2.
$$
Also, a bit later, you're going to need this formula:
$$
sin(2theta)=2sinthetacostheta.
$$
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
$endgroup$
Yes, your solution so far is correct. Now, to integrate $int sin^2theta,dtheta$, use the half-angle formula for the sine function:
$$
sin^2theta=frac1-cos(2theta)2.
$$
Also, a bit later, you're going to need this formula:
$$
sin(2theta)=2sinthetacostheta.
$$
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
edited Feb 3 at 17:15
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
answered Feb 3 at 17:02
Michael RybkinMichael Rybkin
2,833416
2,833416
add a comment |
add a comment |
$begingroup$
$$I=intfracsqrtx^2-9x^3dx=intfrac3sqrt(x/3)^2-1x^3dx$$
and we know that $cosh^2theta-1=sinh^2theta$
$$x=3cosh(y)$$
$$dx=3sinh(y)dy$$
$$I=9intfracsqrtcosh^2(y)-127cosh^3(y)sinh(y)dy=frac13intfracsinh^2(y)cosh^3(y)dy=frac13inttextsech(y)-textsech^3(y)dy$$
a reduction formula can then be used
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
$$I=intfracsqrtx^2-9x^3dx=intfrac3sqrt(x/3)^2-1x^3dx$$
and we know that $cosh^2theta-1=sinh^2theta$
$$x=3cosh(y)$$
$$dx=3sinh(y)dy$$
$$I=9intfracsqrtcosh^2(y)-127cosh^3(y)sinh(y)dy=frac13intfracsinh^2(y)cosh^3(y)dy=frac13inttextsech(y)-textsech^3(y)dy$$
a reduction formula can then be used
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
$$I=intfracsqrtx^2-9x^3dx=intfrac3sqrt(x/3)^2-1x^3dx$$
and we know that $cosh^2theta-1=sinh^2theta$
$$x=3cosh(y)$$
$$dx=3sinh(y)dy$$
$$I=9intfracsqrtcosh^2(y)-127cosh^3(y)sinh(y)dy=frac13intfracsinh^2(y)cosh^3(y)dy=frac13inttextsech(y)-textsech^3(y)dy$$
a reduction formula can then be used
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
$endgroup$
$$I=intfracsqrtx^2-9x^3dx=intfrac3sqrt(x/3)^2-1x^3dx$$
and we know that $cosh^2theta-1=sinh^2theta$
$$x=3cosh(y)$$
$$dx=3sinh(y)dy$$
$$I=9intfracsqrtcosh^2(y)-127cosh^3(y)sinh(y)dy=frac13intfracsinh^2(y)cosh^3(y)dy=frac13inttextsech(y)-textsech^3(y)dy$$
a reduction formula can then be used
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
answered Feb 3 at 20:29
Henry LeeHenry Lee
2,054219
2,054219
add a comment |
add a comment |
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$begingroup$
Yes, you. However, I think a hyperbolic substitution would be a bit shorter.
$endgroup$
– Bernard
Feb 3 at 16:46
1
$begingroup$
On the right track except for some missing $=$ and $dtheta$.
$endgroup$
– Bernard Massé
Feb 3 at 17:06