Integral of $sin^5xcdotcos^14x$ The Next CEO of Stack OverflowEvaluate $int frac1sin xcos x dx $How to solve $int cos(2x)cos(3x) dx$?Integral $int limits _0 ^pi |sin x + cos x|; dx$Integrate $int fracsin(3x)cos x~dx$Finding $int ^pi_-pisin(nx)cos(mx)dx$Going from integral of $dx$ to integral of $dtheta$?Fourier Series for $f(x)=sin(x)+cos(2x)$Evaluate the definite integral: $intlimits_0^pi/2fracsin x-cos xsqrt1-sin 2x, dx$Improper integral of $exp[-(a+bi)x]log x$Integral of $sin^5(x)cos(x)$
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Integral of $sin^5xcdotcos^14x$
The Next CEO of Stack OverflowEvaluate $int frac1sin xcos x dx $How to solve $int cos(2x)cos(3x) dx$?Integral $int limits _0 ^pi |sin x + cos x|; dx$Integrate $int fracsin(3x)cos x~dx$Finding $int ^pi_-pisin(nx)cos(mx)dx$Going from integral of $dx$ to integral of $dtheta$?Fourier Series for $f(x)=sin(x)+cos(2x)$Evaluate the definite integral: $intlimits_0^pi/2fracsin x-cos xsqrt1-sin 2x, dx$Improper integral of $exp[-(a+bi)x]log x$Integral of $sin^5(x)cos(x)$
$begingroup$
I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.
$$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$
My work:
$$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$
$$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$
substitute $cos(x)$ for $u$.
$du=sin(x)dx$
$$intlimits_0^fracpi2(1-u^2)^2u^14du$$
$$intlimits_0^fracpi2(1-2u+u^4)u^14du$$
$$intlimits_0^fracpi2u^14-2u^16+u^18du$$
$$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$
$$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$
$$0-0+0-(1-1+1)$$
$$-1+1-1=-1$$
The correct answer is $frac84845$
Where did I go wrong?
integration definite-integrals
edited Mar 18 at 11:30
MarianD
1,7391617
asked Mar 18 at 10:56
HarukuHaruku
353
$endgroup$
add a comment |
$begingroup$
I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.
$$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$
My work:
$$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$
$$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$
substitute $cos(x)$ for $u$.
$du=sin(x)dx$
$$intlimits_0^fracpi2(1-u^2)^2u^14du$$
$$intlimits_0^fracpi2(1-2u+u^4)u^14du$$
$$intlimits_0^fracpi2u^14-2u^16+u^18du$$
$$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$
$$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$
$$0-0+0-(1-1+1)$$
$$-1+1-1=-1$$
The correct answer is $frac84845$
Where did I go wrong?
integration definite-integrals
edited Mar 18 at 11:30
MarianD
1,7391617
asked Mar 18 at 10:56
HarukuHaruku
353
$endgroup$
add a comment |
$begingroup$
I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.
$$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$
My work:
$$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$
$$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$
substitute $cos(x)$ for $u$.
$du=sin(x)dx$
$$intlimits_0^fracpi2(1-u^2)^2u^14du$$
$$intlimits_0^fracpi2(1-2u+u^4)u^14du$$
$$intlimits_0^fracpi2u^14-2u^16+u^18du$$
$$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$
$$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$
$$0-0+0-(1-1+1)$$
$$-1+1-1=-1$$
The correct answer is $frac84845$
Where did I go wrong?
integration definite-integrals
edited Mar 18 at 11:30
MarianD
1,7391617
asked Mar 18 at 10:56
HarukuHaruku
353
$endgroup$
I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.
$$intlimits_0^fracpi2sin^5(x)cos^14(x)dx$$
My work:
$$intlimits_0^fracpi2[sin^2(x)]^2cos^14(x)sin(x)dx$$
$$intlimits_0^fracpi2[1-cos^2(x)]^2cos^14(x)sin(x)dx$$
substitute $cos(x)$ for $u$.
$du=sin(x)dx$
$$intlimits_0^fracpi2(1-u^2)^2u^14du$$
$$intlimits_0^fracpi2(1-2u+u^4)u^14du$$
$$intlimits_0^fracpi2u^14-2u^16+u^18du$$
$$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=$$
$$frac115cos^15left(fracpi2right)-frac217cos^17left(fracpi2right)+frac119cos^19left(fracpi2right) - left(frac115cos^15(0)-frac217cos^17(0)+frac119cos^19(0)right)$$
$$0-0+0-(1-1+1)$$
$$-1+1-1=-1$$
The correct answer is $frac84845$
Where did I go wrong?
integration definite-integrals
integration definite-integrals
edited Mar 18 at 11:30
MarianD
1,7391617
asked Mar 18 at 10:56
HarukuHaruku
353
edited Mar 18 at 11:30
MarianD
1,7391617
asked Mar 18 at 10:56
HarukuHaruku
353
edited Mar 18 at 11:30
MarianD
1,7391617
edited Mar 18 at 11:30
MarianD
1,7391617
edited Mar 18 at 11:30
MarianD
1,7391617
1,7391617
asked Mar 18 at 10:56
HarukuHaruku
353
asked Mar 18 at 10:56
HarukuHaruku
353
asked Mar 18 at 10:56
HarukuHaruku
353
353
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
add a comment |
$begingroup$
$$u = cos x$$
$$du = -sin x ,dx$$
(and not $du = sin x ,dx$, as you wrote).
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
$endgroup$
add a comment |
$begingroup$
There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
$$u=cos(x)$$
then the limit of integration change to
$$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)LARGE_colorredcos0^colorredcosfracpi2=F(cos(x))LARGE_colorred0^colorredfracpi2$$
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
add a comment |
$begingroup$
It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
add a comment |
$begingroup$
It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
It is in fact $u=-cos x$ and by substitution we get $$frac115u^15-frac217u^17+frac119u^19Bigg|_-1^0=1over 15-2over 17+1over 19=8over 4845$$You where only wrong in substitution...
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 11:02
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
add a comment |
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help.
$endgroup$
– Haruku
Mar 18 at 11:09
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
$begingroup$
Your welcome. Good luck!!
$endgroup$
– Mostafa Ayaz
Mar 18 at 11:10
add a comment |
$begingroup$
$$u = cos x$$
$$du = -sin x ,dx$$
(and not $du = sin x ,dx$, as you wrote).
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
$endgroup$
add a comment |
$begingroup$
$$u = cos x$$
$$du = -sin x ,dx$$
(and not $du = sin x ,dx$, as you wrote).
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
$endgroup$
add a comment |
$begingroup$
$$u = cos x$$
$$du = -sin x ,dx$$
(and not $du = sin x ,dx$, as you wrote).
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
$endgroup$
$$u = cos x$$
$$du = -sin x ,dx$$
(and not $du = sin x ,dx$, as you wrote).
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
answered Mar 18 at 11:05
MarianDMarianD
1,7391617
1,7391617
add a comment |
add a comment |
$begingroup$
There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
$$u=cos(x)$$
then the limit of integration change to
$$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)LARGE_colorredcos0^colorredcosfracpi2=F(cos(x))LARGE_colorred0^colorredfracpi2$$
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
$endgroup$
add a comment |
$begingroup$
There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
$$u=cos(x)$$
then the limit of integration change to
$$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)LARGE_colorredcos0^colorredcosfracpi2=F(cos(x))LARGE_colorred0^colorredfracpi2$$
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
$endgroup$
add a comment |
$begingroup$
There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
$$u=cos(x)$$
then the limit of integration change to
$$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)LARGE_colorredcos0^colorredcosfracpi2=F(cos(x))LARGE_colorred0^colorredfracpi2$$
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
$endgroup$
There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is
$$u=cos(x)$$
then the limit of integration change to
$$intlimits_colorred0^colorredfracpi2f(cos(x))(-sin(x))dx=intlimits_colorredcos0^colorredcosfracpi2f(u)du=F(u)LARGE_colorredcos0^colorredcosfracpi2=F(cos(x))LARGE_colorred0^colorredfracpi2$$
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
edited Mar 20 at 11:44
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
answered Mar 18 at 11:26
miracle173miracle173
7,38022247
7,38022247
add a comment |
add a comment |
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