Integrate$ intfrac sin^8 x - cos^8x 1 - 2sin^2 x cdot cos^2 x mathrmdx$ [on hold]
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
Help with the question.
$$
intfrac sin^8 x - cos^8x
1 - 2sin^2 x cdot cos^2 x mathrmdx
$$
also what I have tried is
$$
intfrac( sin^4 x + cos^4x) cdot ( sin^2 x + cos^2 x) cdot ( sin^2 x - cos^2 x )
1 - 2sin^2x cdot cos^2 x mathrmdx
$$
after this I am stuck
Also I do not want to use
$$
sin^4 x + cos^4 x = (sin^2 x + cos ^2 x )^2 -2sin^2xcdotcos^2x
$$
or
$$1 - 2 sin^2xcdotcos^2x = sin^4 x + cos^4 x $$
I did not get the required answer
integration trigonometry derivatives indefinite-integrals trigonometric-integrals
edited Aug 29 at 18:38
Abcd
2,5581926
asked Aug 27 at 14:59
William Wills
1
put on hold as off-topic by user21820, Holo, Did, Andrés E. Caicedo, amWhy 6 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Holo, Did, amWhy
 |Â
show 7 more comments
up vote
-2
down vote
favorite
Help with the question.
$$
intfrac sin^8 x - cos^8x
1 - 2sin^2 x cdot cos^2 x mathrmdx
$$
also what I have tried is
$$
intfrac( sin^4 x + cos^4x) cdot ( sin^2 x + cos^2 x) cdot ( sin^2 x - cos^2 x )
1 - 2sin^2x cdot cos^2 x mathrmdx
$$
after this I am stuck
Also I do not want to use
$$
sin^4 x + cos^4 x = (sin^2 x + cos ^2 x )^2 -2sin^2xcdotcos^2x
$$
or
$$1 - 2 sin^2xcdotcos^2x = sin^4 x + cos^4 x $$
I did not get the required answer
integration trigonometry derivatives indefinite-integrals trigonometric-integrals
edited Aug 29 at 18:38
Abcd
2,5581926
asked Aug 27 at 14:59
William Wills
1
put on hold as off-topic by user21820, Holo, Did, Andrés E. Caicedo, amWhy 6 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Holo, Did, amWhy
2
Hint: $1 = (sin^2(x)+ cos^2(x))^2 = cdots$
– xbh
Aug 27 at 15:04
1
Was this same question not asked and answered yesterday?
– Sobi
Aug 27 at 15:07
We can use substitution for $tan(x)$
– Rumpelstiltskin
Aug 27 at 15:09
You can use that $sinx^2+cosx^2=1$. Expand the rest and use thatv$sinx^2=1-cosx^2$ and $cosx^2=1-sinx^2$
– Fareed AF
Aug 27 at 15:15
1
You should clarify what kind of answer you are looking for compared to the one already posted on this same question you asked last day. You could have done this by editing that post itself instead of asking it again. This looks like an exact duplicate of your previous post.
– StubbornAtom
Aug 27 at 15:20
 |Â
show 7 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Help with the question.
$$
intfrac sin^8 x - cos^8x
1 - 2sin^2 x cdot cos^2 x mathrmdx
$$
also what I have tried is
$$
intfrac( sin^4 x + cos^4x) cdot ( sin^2 x + cos^2 x) cdot ( sin^2 x - cos^2 x )
1 - 2sin^2x cdot cos^2 x mathrmdx
$$
after this I am stuck
Also I do not want to use
$$
sin^4 x + cos^4 x = (sin^2 x + cos ^2 x )^2 -2sin^2xcdotcos^2x
$$
or
$$1 - 2 sin^2xcdotcos^2x = sin^4 x + cos^4 x $$
I did not get the required answer
integration trigonometry derivatives indefinite-integrals trigonometric-integrals
edited Aug 29 at 18:38
Abcd
2,5581926
asked Aug 27 at 14:59
William Wills
1
Help with the question.
$$
intfrac sin^8 x - cos^8x
1 - 2sin^2 x cdot cos^2 x mathrmdx
$$
also what I have tried is
$$
intfrac( sin^4 x + cos^4x) cdot ( sin^2 x + cos^2 x) cdot ( sin^2 x - cos^2 x )
1 - 2sin^2x cdot cos^2 x mathrmdx
$$
after this I am stuck
Also I do not want to use
$$
sin^4 x + cos^4 x = (sin^2 x + cos ^2 x )^2 -2sin^2xcdotcos^2x
$$
or
$$1 - 2 sin^2xcdotcos^2x = sin^4 x + cos^4 x $$
I did not get the required answer
integration trigonometry derivatives indefinite-integrals trigonometric-integrals
edited Aug 29 at 18:38
Abcd
2,5581926
asked Aug 27 at 14:59
William Wills
1
edited Aug 29 at 18:38
Abcd
2,5581926
edited Aug 29 at 18:38
Abcd
2,5581926
edited Aug 29 at 18:38
Abcd
2,5581926
2,5581926
asked Aug 27 at 14:59
William Wills
1
asked Aug 27 at 14:59
William Wills
1
asked Aug 27 at 14:59
William Wills
1
1
put on hold as off-topic by user21820, Holo, Did, Andrés E. Caicedo, amWhy 6 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Holo, Did, amWhy
put on hold as off-topic by user21820, Holo, Did, Andrés E. Caicedo, amWhy 6 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Holo, Did, amWhy
2
Hint: $1 = (sin^2(x)+ cos^2(x))^2 = cdots$
– xbh
Aug 27 at 15:04
1
Was this same question not asked and answered yesterday?
– Sobi
Aug 27 at 15:07
We can use substitution for $tan(x)$
– Rumpelstiltskin
Aug 27 at 15:09
You can use that $sinx^2+cosx^2=1$. Expand the rest and use thatv$sinx^2=1-cosx^2$ and $cosx^2=1-sinx^2$
– Fareed AF
Aug 27 at 15:15
1
You should clarify what kind of answer you are looking for compared to the one already posted on this same question you asked last day. You could have done this by editing that post itself instead of asking it again. This looks like an exact duplicate of your previous post.
– StubbornAtom
Aug 27 at 15:20
 |Â
show 7 more comments
2
Hint: $1 = (sin^2(x)+ cos^2(x))^2 = cdots$
– xbh
Aug 27 at 15:04
1
Was this same question not asked and answered yesterday?
– Sobi
Aug 27 at 15:07
We can use substitution for $tan(x)$
– Rumpelstiltskin
Aug 27 at 15:09
You can use that $sinx^2+cosx^2=1$. Expand the rest and use thatv$sinx^2=1-cosx^2$ and $cosx^2=1-sinx^2$
– Fareed AF
Aug 27 at 15:15
1
You should clarify what kind of answer you are looking for compared to the one already posted on this same question you asked last day. You could have done this by editing that post itself instead of asking it again. This looks like an exact duplicate of your previous post.
– StubbornAtom
Aug 27 at 15:20
2
2
Hint: $1 = (sin^2(x)+ cos^2(x))^2 = cdots$
– xbh
Aug 27 at 15:04
Hint: $1 = (sin^2(x)+ cos^2(x))^2 = cdots$
– xbh
Aug 27 at 15:04
1
1
Was this same question not asked and answered yesterday?
– Sobi
Aug 27 at 15:07
Was this same question not asked and answered yesterday?
– Sobi
Aug 27 at 15:07
We can use substitution for $tan(x)$
– Rumpelstiltskin
Aug 27 at 15:09
We can use substitution for $tan(x)$
– Rumpelstiltskin
Aug 27 at 15:09
You can use that $sinx^2+cosx^2=1$. Expand the rest and use thatv$sinx^2=1-cosx^2$ and $cosx^2=1-sinx^2$
– Fareed AF
Aug 27 at 15:15
You can use that $sinx^2+cosx^2=1$. Expand the rest and use thatv$sinx^2=1-cosx^2$ and $cosx^2=1-sinx^2$
– Fareed AF
Aug 27 at 15:15
1
1
You should clarify what kind of answer you are looking for compared to the one already posted on this same question you asked last day. You could have done this by editing that post itself instead of asking it again. This looks like an exact duplicate of your previous post.
– StubbornAtom
Aug 27 at 15:20
You should clarify what kind of answer you are looking for compared to the one already posted on this same question you asked last day. You could have done this by editing that post itself instead of asking it again. This looks like an exact duplicate of your previous post.
– StubbornAtom
Aug 27 at 15:20
 |Â
show 7 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
Hint: Your Integrand is equal to $$1-2cos^2(x)$$
For the proof, Show that
$$(1-2cos^2(x))(1-2sin^2(x)cos^2(x))-(sin(x)^8-cos(x)^8)=0$$
It is
$$sin^8(x)-cos^8(x)=(sin^2(x)-cos^2(x))(sin^4(x)+cos^4(x))=$$
$$-cos(2x)(sin^4(x)+cos^4(x))$$
and from
$$sin^2(x)+cos^2(x)=1$$ we get by squaring
$$sin^4(x)+cos^4(x)=1-2sin^2(x)cos^2(x)$$
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
But how did this come ??
– William Wills
Aug 27 at 15:46
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
add a comment |Â
up vote
0
down vote
Hint:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x.$$
answered Aug 27 at 15:38
Bernard
111k635102
add a comment |Â
up vote
0
down vote
Given $$int dfracsin^8x-cos^8x1-2sin^2xcos^2xdx$$Now multiply numerator and denominator by $dfrac4sec^4xcos4x+3$ and we get$$intdfrac(tan^2x-1)sec^2x(1+tan^2x)^2$$
And now use substitution $u=tan x$ and $du=sec^2xdx$and if you find $$intdfracu^2-1(u^2+1)^2du$$ you will get the answer
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Your Integrand is equal to $$1-2cos^2(x)$$
For the proof, Show that
$$(1-2cos^2(x))(1-2sin^2(x)cos^2(x))-(sin(x)^8-cos(x)^8)=0$$
It is
$$sin^8(x)-cos^8(x)=(sin^2(x)-cos^2(x))(sin^4(x)+cos^4(x))=$$
$$-cos(2x)(sin^4(x)+cos^4(x))$$
and from
$$sin^2(x)+cos^2(x)=1$$ we get by squaring
$$sin^4(x)+cos^4(x)=1-2sin^2(x)cos^2(x)$$
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
But how did this come ??
– William Wills
Aug 27 at 15:46
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
add a comment |Â
up vote
1
down vote
Hint: Your Integrand is equal to $$1-2cos^2(x)$$
For the proof, Show that
$$(1-2cos^2(x))(1-2sin^2(x)cos^2(x))-(sin(x)^8-cos(x)^8)=0$$
It is
$$sin^8(x)-cos^8(x)=(sin^2(x)-cos^2(x))(sin^4(x)+cos^4(x))=$$
$$-cos(2x)(sin^4(x)+cos^4(x))$$
and from
$$sin^2(x)+cos^2(x)=1$$ we get by squaring
$$sin^4(x)+cos^4(x)=1-2sin^2(x)cos^2(x)$$
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
But how did this come ??
– William Wills
Aug 27 at 15:46
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Your Integrand is equal to $$1-2cos^2(x)$$
For the proof, Show that
$$(1-2cos^2(x))(1-2sin^2(x)cos^2(x))-(sin(x)^8-cos(x)^8)=0$$
It is
$$sin^8(x)-cos^8(x)=(sin^2(x)-cos^2(x))(sin^4(x)+cos^4(x))=$$
$$-cos(2x)(sin^4(x)+cos^4(x))$$
and from
$$sin^2(x)+cos^2(x)=1$$ we get by squaring
$$sin^4(x)+cos^4(x)=1-2sin^2(x)cos^2(x)$$
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
Hint: Your Integrand is equal to $$1-2cos^2(x)$$
For the proof, Show that
$$(1-2cos^2(x))(1-2sin^2(x)cos^2(x))-(sin(x)^8-cos(x)^8)=0$$
It is
$$sin^8(x)-cos^8(x)=(sin^2(x)-cos^2(x))(sin^4(x)+cos^4(x))=$$
$$-cos(2x)(sin^4(x)+cos^4(x))$$
and from
$$sin^2(x)+cos^2(x)=1$$ we get by squaring
$$sin^4(x)+cos^4(x)=1-2sin^2(x)cos^2(x)$$
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
edited Aug 27 at 16:08
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
answered Aug 27 at 15:15
Dr. Sonnhard Graubner
68.2k32660
68.2k32660
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
But how did this come ??
– William Wills
Aug 27 at 15:46
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
add a comment |Â
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
But how did this come ??
– William Wills
Aug 27 at 15:46
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
I.e. $-cos 2x$?
– Bernard
Aug 27 at 15:34
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
$$-cos(2x)=-(2cos^2(x)-1)$$, yes it is.
– Dr. Sonnhard Graubner
Aug 27 at 15:37
But how did this come ??
– William Wills
Aug 27 at 15:46
But how did this come ??
– William Wills
Aug 27 at 15:46
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
@Dr.SonnhardGraubner could you please tell how to prove (1−2cos2(x))(1−2sin2(x)cos2(x))−(sin(x)8−cos(x)8)=0
– William Wills
Aug 27 at 15:57
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
Very neat trick, Dr. Graubner!
– bjcolby15
Aug 27 at 22:38
add a comment |Â
up vote
0
down vote
Hint:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x.$$
answered Aug 27 at 15:38
Bernard
111k635102
add a comment |Â
up vote
0
down vote
Hint:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x.$$
answered Aug 27 at 15:38
Bernard
111k635102
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x.$$
answered Aug 27 at 15:38
Bernard
111k635102
Hint:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x.$$
answered Aug 27 at 15:38
Bernard
111k635102
answered Aug 27 at 15:38
Bernard
111k635102
answered Aug 27 at 15:38
Bernard
111k635102
answered Aug 27 at 15:38
Bernard
111k635102
111k635102
add a comment |Â
add a comment |Â
up vote
0
down vote
Given $$int dfracsin^8x-cos^8x1-2sin^2xcos^2xdx$$Now multiply numerator and denominator by $dfrac4sec^4xcos4x+3$ and we get$$intdfrac(tan^2x-1)sec^2x(1+tan^2x)^2$$
And now use substitution $u=tan x$ and $du=sec^2xdx$and if you find $$intdfracu^2-1(u^2+1)^2du$$ you will get the answer
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
add a comment |Â
up vote
0
down vote
Given $$int dfracsin^8x-cos^8x1-2sin^2xcos^2xdx$$Now multiply numerator and denominator by $dfrac4sec^4xcos4x+3$ and we get$$intdfrac(tan^2x-1)sec^2x(1+tan^2x)^2$$
And now use substitution $u=tan x$ and $du=sec^2xdx$and if you find $$intdfracu^2-1(u^2+1)^2du$$ you will get the answer
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Given $$int dfracsin^8x-cos^8x1-2sin^2xcos^2xdx$$Now multiply numerator and denominator by $dfrac4sec^4xcos4x+3$ and we get$$intdfrac(tan^2x-1)sec^2x(1+tan^2x)^2$$
And now use substitution $u=tan x$ and $du=sec^2xdx$and if you find $$intdfracu^2-1(u^2+1)^2du$$ you will get the answer
Given $$int dfracsin^8x-cos^8x1-2sin^2xcos^2xdx$$Now multiply numerator and denominator by $dfrac4sec^4xcos4x+3$ and we get$$intdfrac(tan^2x-1)sec^2x(1+tan^2x)^2$$
And now use substitution $u=tan x$ and $du=sec^2xdx$and if you find $$intdfracu^2-1(u^2+1)^2du$$ you will get the answer
answered Aug 27 at 15:43
user587574
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
add a comment |Â
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
How did you think of second step that is Now multiply numerator and denominator by 4sec4xcos4x+3 ?? And also how did you get the expression in third step ?
– William Wills
Aug 27 at 16:03
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@WilliamWills For such type of integrals, it would be easy to bring it in the form of $tan$ and then use substitution. So, eventually I did.
– user587574
Aug 28 at 18:09
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
@user587574 but how did you convert integrand into tan x . how did you think of multiplying by 4sec4x(cos4x+3) ??
– William Wills
Aug 29 at 1:03
add a comment |Â
2
Hint: $1 = (sin^2(x)+ cos^2(x))^2 = cdots$
– xbh
Aug 27 at 15:04
1
Was this same question not asked and answered yesterday?
– Sobi
Aug 27 at 15:07
We can use substitution for $tan(x)$
– Rumpelstiltskin
Aug 27 at 15:09
You can use that $sinx^2+cosx^2=1$. Expand the rest and use thatv$sinx^2=1-cosx^2$ and $cosx^2=1-sinx^2$
– Fareed AF
Aug 27 at 15:15
1
You should clarify what kind of answer you are looking for compared to the one already posted on this same question you asked last day. You could have done this by editing that post itself instead of asking it again. This looks like an exact duplicate of your previous post.
– StubbornAtom
Aug 27 at 15:20