How to integrate $4cos^2 (theta) sin^2 (theta)$
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Should I change it into $4 cos^2 (theta) (1-cos^2(theta))$ or is there any possible way to use product rule?
integration trigonometry
edited Dec 14 '18 at 14:04
Namaste
1
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
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add a comment |
$begingroup$
Should I change it into $4 cos^2 (theta) (1-cos^2(theta))$ or is there any possible way to use product rule?
integration trigonometry
edited Dec 14 '18 at 14:04
Namaste
1
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
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4
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Think of $sin 2 theta$.
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– preferred_anon
Dec 14 '18 at 13:54
7
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You have been around for 5 months. Haven't you noticed yet that you are supposed to use MathJax in your questions?
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– José Carlos Santos
Dec 14 '18 at 13:55
add a comment |
$begingroup$
Should I change it into $4 cos^2 (theta) (1-cos^2(theta))$ or is there any possible way to use product rule?
integration trigonometry
edited Dec 14 '18 at 14:04
Namaste
1
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
$endgroup$
Should I change it into $4 cos^2 (theta) (1-cos^2(theta))$ or is there any possible way to use product rule?
integration trigonometry
integration trigonometry
edited Dec 14 '18 at 14:04
Namaste
1
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
edited Dec 14 '18 at 14:04
Namaste
1
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
edited Dec 14 '18 at 14:04
Namaste
1
edited Dec 14 '18 at 14:04
Namaste
1
edited Dec 14 '18 at 14:04
Namaste
1
1
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
asked Dec 14 '18 at 13:52
WangcincayWangcincay
413
413
4
$begingroup$
Think of $sin 2 theta$.
$endgroup$
– preferred_anon
Dec 14 '18 at 13:54
7
$begingroup$
You have been around for 5 months. Haven't you noticed yet that you are supposed to use MathJax in your questions?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:55
add a comment |
4
$begingroup$
Think of $sin 2 theta$.
$endgroup$
– preferred_anon
Dec 14 '18 at 13:54
7
$begingroup$
You have been around for 5 months. Haven't you noticed yet that you are supposed to use MathJax in your questions?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:55
4
4
$begingroup$
Think of $sin 2 theta$.
$endgroup$
– preferred_anon
Dec 14 '18 at 13:54
$begingroup$
Think of $sin 2 theta$.
$endgroup$
– preferred_anon
Dec 14 '18 at 13:54
7
7
$begingroup$
You have been around for 5 months. Haven't you noticed yet that you are supposed to use MathJax in your questions?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:55
$begingroup$
You have been around for 5 months. Haven't you noticed yet that you are supposed to use MathJax in your questions?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:55
add a comment |
3 Answers
3
active
oldest
votes
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Hint: Use that $$4sin^2(x)cos^2(x)= (sin(2x))^2$$
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
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$$ int { 4cos ^{ 2 }{ x } sin ^{ 2 }{ x } dx } =int { { sin ^{ 2 }{ left( 2x right) dx } } } \ int { frac { 1-cos { left( 4x right) } }{ 2 } } dx=frac { x }{ 2 } -frac { sin { left( 4x right) } }{ 8 } +c $$
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
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add a comment |
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TIP$_1$: $4sin^2theta cos^2theta = (2sinthetacostheta)^2 = sin^2,2theta$.
TIP$_2$: $cos,2theta = cos^2 theta - sin^2theta = 1 - 2sin^2theta$, for all $theta in Bbb R$.
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Use that $$4sin^2(x)cos^2(x)= (sin(2x))^2$$
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$4sin^2(x)cos^2(x)= (sin(2x))^2$$
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$4sin^2(x)cos^2(x)= (sin(2x))^2$$
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
Hint: Use that $$4sin^2(x)cos^2(x)= (sin(2x))^2$$
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Dec 14 '18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
78.8k42867
add a comment |
add a comment |
$begingroup$
$$ int { 4cos ^{ 2 }{ x } sin ^{ 2 }{ x } dx } =int { { sin ^{ 2 }{ left( 2x right) dx } } } \ int { frac { 1-cos { left( 4x right) } }{ 2 } } dx=frac { x }{ 2 } -frac { sin { left( 4x right) } }{ 8 } +c $$
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
$endgroup$
add a comment |
$begingroup$
$$ int { 4cos ^{ 2 }{ x } sin ^{ 2 }{ x } dx } =int { { sin ^{ 2 }{ left( 2x right) dx } } } \ int { frac { 1-cos { left( 4x right) } }{ 2 } } dx=frac { x }{ 2 } -frac { sin { left( 4x right) } }{ 8 } +c $$
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
$endgroup$
add a comment |
$begingroup$
$$ int { 4cos ^{ 2 }{ x } sin ^{ 2 }{ x } dx } =int { { sin ^{ 2 }{ left( 2x right) dx } } } \ int { frac { 1-cos { left( 4x right) } }{ 2 } } dx=frac { x }{ 2 } -frac { sin { left( 4x right) } }{ 8 } +c $$
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
$endgroup$
$$ int { 4cos ^{ 2 }{ x } sin ^{ 2 }{ x } dx } =int { { sin ^{ 2 }{ left( 2x right) dx } } } \ int { frac { 1-cos { left( 4x right) } }{ 2 } } dx=frac { x }{ 2 } -frac { sin { left( 4x right) } }{ 8 } +c $$
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
answered Dec 14 '18 at 15:56
Lakshya SinhaLakshya Sinha
724
724
add a comment |
add a comment |
$begingroup$
TIP$_1$: $4sin^2theta cos^2theta = (2sinthetacostheta)^2 = sin^2,2theta$.
TIP$_2$: $cos,2theta = cos^2 theta - sin^2theta = 1 - 2sin^2theta$, for all $theta in Bbb R$.
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
TIP$_1$: $4sin^2theta cos^2theta = (2sinthetacostheta)^2 = sin^2,2theta$.
TIP$_2$: $cos,2theta = cos^2 theta - sin^2theta = 1 - 2sin^2theta$, for all $theta in Bbb R$.
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
TIP$_1$: $4sin^2theta cos^2theta = (2sinthetacostheta)^2 = sin^2,2theta$.
TIP$_2$: $cos,2theta = cos^2 theta - sin^2theta = 1 - 2sin^2theta$, for all $theta in Bbb R$.
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
TIP$_1$: $4sin^2theta cos^2theta = (2sinthetacostheta)^2 = sin^2,2theta$.
TIP$_2$: $cos,2theta = cos^2 theta - sin^2theta = 1 - 2sin^2theta$, for all $theta in Bbb R$.
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
answered Dec 14 '18 at 13:59
Lucas HenriqueLucas Henrique
1,031414
1,031414
add a comment |
add a comment |
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4
$begingroup$
Think of $sin 2 theta$.
$endgroup$
– preferred_anon
Dec 14 '18 at 13:54
7
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You have been around for 5 months. Haven't you noticed yet that you are supposed to use MathJax in your questions?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:55