What is $int sqrtcos(2Q) / sin(Q) ,mathrmdQ$? [closed]
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What is
$$
int fracsqrtcos(2Q)sin(Q) ,mathrmdQ?
$$
I have tried all the method which is possible but could not able to find the solution. Can anyone please tell me the solution of this problem.
calculus integration indefinite-integrals
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:29
Rajesh Gupta
92
closed as off-topic by Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500 Aug 28 at 9:12
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." – Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500
 |Â
show 1 more comment
up vote
1
down vote
favorite
What is
$$
int fracsqrtcos(2Q)sin(Q) ,mathrmdQ?
$$
I have tried all the method which is possible but could not able to find the solution. Can anyone please tell me the solution of this problem.
calculus integration indefinite-integrals
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:29
Rajesh Gupta
92
closed as off-topic by Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500 Aug 28 at 9:12
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." – Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500
4
Is this what you meant: $$int fracsqrtcos 2qsin q , mathrmdq$$
– Tolaso
Aug 27 at 12:30
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 27 at 12:31
1
Wolfram Alpha looks rather unpleasant
– gt6989b
Aug 27 at 12:36
Tip: expand the double-angle cosine using the double angle formula.
– FGSUZ
Aug 27 at 13:05
I done that still can't solve
– Rajesh Gupta
Aug 27 at 13:09
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is
$$
int fracsqrtcos(2Q)sin(Q) ,mathrmdQ?
$$
I have tried all the method which is possible but could not able to find the solution. Can anyone please tell me the solution of this problem.
calculus integration indefinite-integrals
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:29
Rajesh Gupta
92
What is
$$
int fracsqrtcos(2Q)sin(Q) ,mathrmdQ?
$$
I have tried all the method which is possible but could not able to find the solution. Can anyone please tell me the solution of this problem.
calculus integration indefinite-integrals
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
asked Aug 27 at 12:29
Rajesh Gupta
92
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
edited Aug 27 at 13:36
Jendrik Stelzner
7,63121037
7,63121037
asked Aug 27 at 12:29
Rajesh Gupta
92
asked Aug 27 at 12:29
Rajesh Gupta
92
asked Aug 27 at 12:29
Rajesh Gupta
92
92
closed as off-topic by Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500 Aug 28 at 9:12
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." – Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500
closed as off-topic by Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500 Aug 28 at 9:12
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." – Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500
4
Is this what you meant: $$int fracsqrtcos 2qsin q , mathrmdq$$
– Tolaso
Aug 27 at 12:30
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 27 at 12:31
1
Wolfram Alpha looks rather unpleasant
– gt6989b
Aug 27 at 12:36
Tip: expand the double-angle cosine using the double angle formula.
– FGSUZ
Aug 27 at 13:05
I done that still can't solve
– Rajesh Gupta
Aug 27 at 13:09
 |Â
show 1 more comment
4
Is this what you meant: $$int fracsqrtcos 2qsin q , mathrmdq$$
– Tolaso
Aug 27 at 12:30
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 27 at 12:31
1
Wolfram Alpha looks rather unpleasant
– gt6989b
Aug 27 at 12:36
Tip: expand the double-angle cosine using the double angle formula.
– FGSUZ
Aug 27 at 13:05
I done that still can't solve
– Rajesh Gupta
Aug 27 at 13:09
4
4
Is this what you meant: $$int fracsqrtcos 2qsin q , mathrmdq$$
– Tolaso
Aug 27 at 12:30
Is this what you meant: $$int fracsqrtcos 2qsin q , mathrmdq$$
– Tolaso
Aug 27 at 12:30
1
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 27 at 12:31
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 27 at 12:31
1
1
Wolfram Alpha looks rather unpleasant
– gt6989b
Aug 27 at 12:36
Wolfram Alpha looks rather unpleasant
– gt6989b
Aug 27 at 12:36
Tip: expand the double-angle cosine using the double angle formula.
– FGSUZ
Aug 27 at 13:05
Tip: expand the double-angle cosine using the double angle formula.
– FGSUZ
Aug 27 at 13:05
I done that still can't solve
– Rajesh Gupta
Aug 27 at 13:09
I done that still can't solve
– Rajesh Gupta
Aug 27 at 13:09
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
First notice that
beginequation
cosleft(2xright)=cos^2left(xright)-sin^2left(xright)
endequation
and
beginequation
sinleft(xright)=dfractanleft(xright)secleft(xright)
endequation
so
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleint
sec^2(x) cdotdfracsqrt1-tan^2left(xright)tanleft(xright)left(tan^2left(xright)+1right),mathrmdx
endequation
Change of variable as
beginequation
u = tan (x)
endequation
we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracsqrt1-u^2uleft(u^2+1right),mathrmdu
endequation
Another change of variable as
beginequation
v = sqrt1 - u^2
endequation
gives us
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-displaystyleintdfracv^2left(v^2-2right)left(v^2-1right),mathrmdv
endequation
Now let's factor the denominator as
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracv^2left(v-1right)left(v+1right)left(v^2-2right),mathrmdv
endequation
Then perform partial fraction decomposition
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintleft(dfrac2v^2-2+dfrac12left(v+1right)-dfrac12left(v-1right)right)mathrmdv
=
2A + frac12B - frac12 C
endequation
Let's do $A$,
beginequation
A = displaystyleintdfrac1v^2-2,mathrmdv
=
=displaystyleintdfrac1left(v-sqrt2right)left(v+sqrt2right),mathrmdv
=displaystyleintleft(dfrac12^frac32left(v-sqrt2right)-dfrac12^frac32left(v+sqrt2right)right)mathrmdv
endequation
which s
beginequation
A
=
=dfraclnleft(v-sqrt2right)2^frac32-dfraclnleft(v+sqrt2right)2^frac32
endequation
Now, similarly
beginequation
B =lnleft(v+1right)
endequation
and
beginequation
C =lnleft(v-1right)
endequation
Plugging all $A,B,C$ back we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-dfraclnleft(v+sqrt2right)sqrt2+dfraclnleft(v-sqrt2right)sqrt2+dfraclnleft(v+1right)2-dfraclnleft(v-1right)2
endequation
Undoing the change of variable $v = sqrt1 - u^2$, we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx=dfraclnleft(sqrt1-u^2+sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2-sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2+1right)2+dfraclnleft(sqrt1-u^2-1right)2
endequation
Undoing the other change of variable $u = tan (x)$, we get
beginequation
dfraclnleft(sqrt1-tan^2left(xright)+sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)-sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)+1right)2+dfraclnleft(sqrt1-tan^2left(xright)-1right)2
endequation
Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First notice that
beginequation
cosleft(2xright)=cos^2left(xright)-sin^2left(xright)
endequation
and
beginequation
sinleft(xright)=dfractanleft(xright)secleft(xright)
endequation
so
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleint
sec^2(x) cdotdfracsqrt1-tan^2left(xright)tanleft(xright)left(tan^2left(xright)+1right),mathrmdx
endequation
Change of variable as
beginequation
u = tan (x)
endequation
we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracsqrt1-u^2uleft(u^2+1right),mathrmdu
endequation
Another change of variable as
beginequation
v = sqrt1 - u^2
endequation
gives us
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-displaystyleintdfracv^2left(v^2-2right)left(v^2-1right),mathrmdv
endequation
Now let's factor the denominator as
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracv^2left(v-1right)left(v+1right)left(v^2-2right),mathrmdv
endequation
Then perform partial fraction decomposition
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintleft(dfrac2v^2-2+dfrac12left(v+1right)-dfrac12left(v-1right)right)mathrmdv
=
2A + frac12B - frac12 C
endequation
Let's do $A$,
beginequation
A = displaystyleintdfrac1v^2-2,mathrmdv
=
=displaystyleintdfrac1left(v-sqrt2right)left(v+sqrt2right),mathrmdv
=displaystyleintleft(dfrac12^frac32left(v-sqrt2right)-dfrac12^frac32left(v+sqrt2right)right)mathrmdv
endequation
which s
beginequation
A
=
=dfraclnleft(v-sqrt2right)2^frac32-dfraclnleft(v+sqrt2right)2^frac32
endequation
Now, similarly
beginequation
B =lnleft(v+1right)
endequation
and
beginequation
C =lnleft(v-1right)
endequation
Plugging all $A,B,C$ back we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-dfraclnleft(v+sqrt2right)sqrt2+dfraclnleft(v-sqrt2right)sqrt2+dfraclnleft(v+1right)2-dfraclnleft(v-1right)2
endequation
Undoing the change of variable $v = sqrt1 - u^2$, we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx=dfraclnleft(sqrt1-u^2+sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2-sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2+1right)2+dfraclnleft(sqrt1-u^2-1right)2
endequation
Undoing the other change of variable $u = tan (x)$, we get
beginequation
dfraclnleft(sqrt1-tan^2left(xright)+sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)-sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)+1right)2+dfraclnleft(sqrt1-tan^2left(xright)-1right)2
endequation
Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
add a comment |Â
up vote
1
down vote
First notice that
beginequation
cosleft(2xright)=cos^2left(xright)-sin^2left(xright)
endequation
and
beginequation
sinleft(xright)=dfractanleft(xright)secleft(xright)
endequation
so
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleint
sec^2(x) cdotdfracsqrt1-tan^2left(xright)tanleft(xright)left(tan^2left(xright)+1right),mathrmdx
endequation
Change of variable as
beginequation
u = tan (x)
endequation
we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracsqrt1-u^2uleft(u^2+1right),mathrmdu
endequation
Another change of variable as
beginequation
v = sqrt1 - u^2
endequation
gives us
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-displaystyleintdfracv^2left(v^2-2right)left(v^2-1right),mathrmdv
endequation
Now let's factor the denominator as
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracv^2left(v-1right)left(v+1right)left(v^2-2right),mathrmdv
endequation
Then perform partial fraction decomposition
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintleft(dfrac2v^2-2+dfrac12left(v+1right)-dfrac12left(v-1right)right)mathrmdv
=
2A + frac12B - frac12 C
endequation
Let's do $A$,
beginequation
A = displaystyleintdfrac1v^2-2,mathrmdv
=
=displaystyleintdfrac1left(v-sqrt2right)left(v+sqrt2right),mathrmdv
=displaystyleintleft(dfrac12^frac32left(v-sqrt2right)-dfrac12^frac32left(v+sqrt2right)right)mathrmdv
endequation
which s
beginequation
A
=
=dfraclnleft(v-sqrt2right)2^frac32-dfraclnleft(v+sqrt2right)2^frac32
endequation
Now, similarly
beginequation
B =lnleft(v+1right)
endequation
and
beginequation
C =lnleft(v-1right)
endequation
Plugging all $A,B,C$ back we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-dfraclnleft(v+sqrt2right)sqrt2+dfraclnleft(v-sqrt2right)sqrt2+dfraclnleft(v+1right)2-dfraclnleft(v-1right)2
endequation
Undoing the change of variable $v = sqrt1 - u^2$, we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx=dfraclnleft(sqrt1-u^2+sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2-sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2+1right)2+dfraclnleft(sqrt1-u^2-1right)2
endequation
Undoing the other change of variable $u = tan (x)$, we get
beginequation
dfraclnleft(sqrt1-tan^2left(xright)+sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)-sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)+1right)2+dfraclnleft(sqrt1-tan^2left(xright)-1right)2
endequation
Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First notice that
beginequation
cosleft(2xright)=cos^2left(xright)-sin^2left(xright)
endequation
and
beginequation
sinleft(xright)=dfractanleft(xright)secleft(xright)
endequation
so
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleint
sec^2(x) cdotdfracsqrt1-tan^2left(xright)tanleft(xright)left(tan^2left(xright)+1right),mathrmdx
endequation
Change of variable as
beginequation
u = tan (x)
endequation
we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracsqrt1-u^2uleft(u^2+1right),mathrmdu
endequation
Another change of variable as
beginequation
v = sqrt1 - u^2
endequation
gives us
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-displaystyleintdfracv^2left(v^2-2right)left(v^2-1right),mathrmdv
endequation
Now let's factor the denominator as
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracv^2left(v-1right)left(v+1right)left(v^2-2right),mathrmdv
endequation
Then perform partial fraction decomposition
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintleft(dfrac2v^2-2+dfrac12left(v+1right)-dfrac12left(v-1right)right)mathrmdv
=
2A + frac12B - frac12 C
endequation
Let's do $A$,
beginequation
A = displaystyleintdfrac1v^2-2,mathrmdv
=
=displaystyleintdfrac1left(v-sqrt2right)left(v+sqrt2right),mathrmdv
=displaystyleintleft(dfrac12^frac32left(v-sqrt2right)-dfrac12^frac32left(v+sqrt2right)right)mathrmdv
endequation
which s
beginequation
A
=
=dfraclnleft(v-sqrt2right)2^frac32-dfraclnleft(v+sqrt2right)2^frac32
endequation
Now, similarly
beginequation
B =lnleft(v+1right)
endequation
and
beginequation
C =lnleft(v-1right)
endequation
Plugging all $A,B,C$ back we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-dfraclnleft(v+sqrt2right)sqrt2+dfraclnleft(v-sqrt2right)sqrt2+dfraclnleft(v+1right)2-dfraclnleft(v-1right)2
endequation
Undoing the change of variable $v = sqrt1 - u^2$, we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx=dfraclnleft(sqrt1-u^2+sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2-sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2+1right)2+dfraclnleft(sqrt1-u^2-1right)2
endequation
Undoing the other change of variable $u = tan (x)$, we get
beginequation
dfraclnleft(sqrt1-tan^2left(xright)+sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)-sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)+1right)2+dfraclnleft(sqrt1-tan^2left(xright)-1right)2
endequation
Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
First notice that
beginequation
cosleft(2xright)=cos^2left(xright)-sin^2left(xright)
endequation
and
beginequation
sinleft(xright)=dfractanleft(xright)secleft(xright)
endequation
so
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleint
sec^2(x) cdotdfracsqrt1-tan^2left(xright)tanleft(xright)left(tan^2left(xright)+1right),mathrmdx
endequation
Change of variable as
beginequation
u = tan (x)
endequation
we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracsqrt1-u^2uleft(u^2+1right),mathrmdu
endequation
Another change of variable as
beginequation
v = sqrt1 - u^2
endequation
gives us
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-displaystyleintdfracv^2left(v^2-2right)left(v^2-1right),mathrmdv
endequation
Now let's factor the denominator as
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintdfracv^2left(v-1right)left(v+1right)left(v^2-2right),mathrmdv
endequation
Then perform partial fraction decomposition
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=displaystyleintleft(dfrac2v^2-2+dfrac12left(v+1right)-dfrac12left(v-1right)right)mathrmdv
=
2A + frac12B - frac12 C
endequation
Let's do $A$,
beginequation
A = displaystyleintdfrac1v^2-2,mathrmdv
=
=displaystyleintdfrac1left(v-sqrt2right)left(v+sqrt2right),mathrmdv
=displaystyleintleft(dfrac12^frac32left(v-sqrt2right)-dfrac12^frac32left(v+sqrt2right)right)mathrmdv
endequation
which s
beginequation
A
=
=dfraclnleft(v-sqrt2right)2^frac32-dfraclnleft(v+sqrt2right)2^frac32
endequation
Now, similarly
beginequation
B =lnleft(v+1right)
endequation
and
beginequation
C =lnleft(v-1right)
endequation
Plugging all $A,B,C$ back we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx
=-dfraclnleft(v+sqrt2right)sqrt2+dfraclnleft(v-sqrt2right)sqrt2+dfraclnleft(v+1right)2-dfraclnleft(v-1right)2
endequation
Undoing the change of variable $v = sqrt1 - u^2$, we get
beginequation
displaystyleintdfracsqrtcosleft(2xright)sinleft(xright),mathrmdx=dfraclnleft(sqrt1-u^2+sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2-sqrt2right)sqrt2-dfraclnleft(sqrt1-u^2+1right)2+dfraclnleft(sqrt1-u^2-1right)2
endequation
Undoing the other change of variable $u = tan (x)$, we get
beginequation
dfraclnleft(sqrt1-tan^2left(xright)+sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)-sqrt2right)sqrt2-dfraclnleft(sqrt1-tan^2left(xright)+1right)2+dfraclnleft(sqrt1-tan^2left(xright)-1right)2
endequation
Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
edited Aug 27 at 13:45
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
answered Aug 27 at 13:39
Ahmad Bazzi
4,2861622
4,2861622
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
add a comment |Â
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
what is the mistake @RajeshGupta?
– Ahmad Bazzi
Aug 27 at 15:51
add a comment |Â
4
Is this what you meant: $$int fracsqrtcos 2qsin q , mathrmdq$$
– Tolaso
Aug 27 at 12:30
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 27 at 12:31
1
Wolfram Alpha looks rather unpleasant
– gt6989b
Aug 27 at 12:36
Tip: expand the double-angle cosine using the double angle formula.
– FGSUZ
Aug 27 at 13:05
I done that still can't solve
– Rajesh Gupta
Aug 27 at 13:09