How to solve $int sin^3(x) cos^2(x) dx$ with integration by parts?
Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried
$$int sin^3(x) cos^2(x)dx$$
Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$
$$int v^3 u'^2$$
And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents
calculus real-analysis integration trigonometry
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
asked Jan 31 '18 at 20:11
TreyTrey
304113
add a comment |
Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried
$$int sin^3(x) cos^2(x)dx$$
Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$
$$int v^3 u'^2$$
And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents
calculus real-analysis integration trigonometry
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
asked Jan 31 '18 at 20:11
TreyTrey
304113
2
dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16
add a comment |
Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried
$$int sin^3(x) cos^2(x)dx$$
Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$
$$int v^3 u'^2$$
And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents
calculus real-analysis integration trigonometry
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
asked Jan 31 '18 at 20:11
TreyTrey
304113
Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried
$$int sin^3(x) cos^2(x)dx$$
Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$
$$int v^3 u'^2$$
And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents
calculus real-analysis integration trigonometry
calculus real-analysis integration trigonometry
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
asked Jan 31 '18 at 20:11
TreyTrey
304113
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
asked Jan 31 '18 at 20:11
TreyTrey
304113
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
edited Jan 31 '18 at 20:25
Michael Rozenberg
97.4k1589188
97.4k1589188
asked Jan 31 '18 at 20:11
TreyTrey
304113
asked Jan 31 '18 at 20:11
TreyTrey
304113
asked Jan 31 '18 at 20:11
TreyTrey
304113
304113
2
dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16
add a comment |
2
dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16
2
2
dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16
dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16
add a comment |
5 Answers
5
active
oldest
votes
split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
add a comment |
$$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
add a comment |
$$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
$$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
$$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
add a comment |
Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.
$int sin^3x cos^2 x dx$
pick something easy to integrate to be $v'$
$v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.
lets go with the first one.
$u = sin^2 x, dv = sin xcos^2 x dx\
du = 2sin xcos x dx, v = -frac 13 cos^3x$
$-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
-frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
frac 15cos^5 x - frac 13 cos^3 x$
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
add a comment |
Integration by parts is unnecessary. Do this
$$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
add a comment |
split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
add a comment |
split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Jan 31 '18 at 20:14
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
73.5k42865
add a comment |
add a comment |
$$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
add a comment |
$$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
add a comment |
$$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
$$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
answered Jan 31 '18 at 20:15
haqnaturalhaqnatural
20.6k72457
20.6k72457
add a comment |
add a comment |
$$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
$$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
$$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
add a comment |
$$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
$$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
$$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
add a comment |
$$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
$$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
$$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
$$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
$$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
$$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
answered Jan 31 '18 at 20:17
Michael RozenbergMichael Rozenberg
97.4k1589188
97.4k1589188
add a comment |
add a comment |
Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.
$int sin^3x cos^2 x dx$
pick something easy to integrate to be $v'$
$v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.
lets go with the first one.
$u = sin^2 x, dv = sin xcos^2 x dx\
du = 2sin xcos x dx, v = -frac 13 cos^3x$
$-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
-frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
frac 15cos^5 x - frac 13 cos^3 x$
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
add a comment |
Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.
$int sin^3x cos^2 x dx$
pick something easy to integrate to be $v'$
$v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.
lets go with the first one.
$u = sin^2 x, dv = sin xcos^2 x dx\
du = 2sin xcos x dx, v = -frac 13 cos^3x$
$-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
-frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
frac 15cos^5 x - frac 13 cos^3 x$
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
add a comment |
Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.
$int sin^3x cos^2 x dx$
pick something easy to integrate to be $v'$
$v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.
lets go with the first one.
$u = sin^2 x, dv = sin xcos^2 x dx\
du = 2sin xcos x dx, v = -frac 13 cos^3x$
$-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
-frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
frac 15cos^5 x - frac 13 cos^3 x$
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.
$int sin^3x cos^2 x dx$
pick something easy to integrate to be $v'$
$v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.
lets go with the first one.
$u = sin^2 x, dv = sin xcos^2 x dx\
du = 2sin xcos x dx, v = -frac 13 cos^3x$
$-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
-frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
frac 15cos^5 x - frac 13 cos^3 x$
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
answered Dec 12 '18 at 22:09
Doug MDoug M
44.2k31854
44.2k31854
add a comment |
add a comment |
Integration by parts is unnecessary. Do this
$$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
add a comment |
Integration by parts is unnecessary. Do this
$$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
add a comment |
Integration by parts is unnecessary. Do this
$$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
Integration by parts is unnecessary. Do this
$$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
answered Dec 12 '18 at 21:52
ncmathsadistncmathsadist
42.4k259102
42.4k259102
add a comment |
add a comment |
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dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16