indefinite integration of polynomial and trignometric functionsindefinite integral of $x^nsin(x)$Integrate...
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indefinite integration of polynomial and trignometric functions
indefinite integral of $x^nsin(x)$Integrate $int_0^pi{{xsin x}over{1+cos^2x}}dx$.Solve $int 3xcos(2x)dx$ with integration by partsintegration by parts of trig functionsIntegration techniques for $int x^3sin x^2,dx$Integrations by partsEvaluate $int e^{2cos x}cos 2x dx$How to solve this integration problem by parts and substitution?Indefinite Integration of a trig functionIntegral of $int_0^b cos(x)cos(frac a x)dx$
$begingroup$
integrate $$int frac{(1+ x^2)(2+ x^2)}{(x cos x + sin x)^4}dx$$
I have tried integration by parts but this x cos x + sin x part is creating problem so i tried substituting it, but unlike x sin x + cos x it is not reducing terms. I think something would be multiplied and divided in the first step, though I am not able to figure out what it is.
integration
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
asked Mar 20 at 18:48
user185442user185442
395
$endgroup$
add a comment |
$begingroup$
integrate $$int frac{(1+ x^2)(2+ x^2)}{(x cos x + sin x)^4}dx$$
I have tried integration by parts but this x cos x + sin x part is creating problem so i tried substituting it, but unlike x sin x + cos x it is not reducing terms. I think something would be multiplied and divided in the first step, though I am not able to figure out what it is.
integration
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
asked Mar 20 at 18:48
user185442user185442
395
$endgroup$
$begingroup$
Is this right so?
$endgroup$
– Dr. Sonnhard Graubner
Mar 20 at 18:55
$begingroup$
yes sir, it is and answer is ((x tan x -1)/(x + tan x)) +(1/3)*(((x tan x -1)^3)/(x + tan x)) + c but i do not know the solution
$endgroup$
– user185442
Mar 21 at 17:42
add a comment |
$begingroup$
integrate $$int frac{(1+ x^2)(2+ x^2)}{(x cos x + sin x)^4}dx$$
I have tried integration by parts but this x cos x + sin x part is creating problem so i tried substituting it, but unlike x sin x + cos x it is not reducing terms. I think something would be multiplied and divided in the first step, though I am not able to figure out what it is.
integration
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
asked Mar 20 at 18:48
user185442user185442
395
$endgroup$
integrate $$int frac{(1+ x^2)(2+ x^2)}{(x cos x + sin x)^4}dx$$
I have tried integration by parts but this x cos x + sin x part is creating problem so i tried substituting it, but unlike x sin x + cos x it is not reducing terms. I think something would be multiplied and divided in the first step, though I am not able to figure out what it is.
integration
integration
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
asked Mar 20 at 18:48
user185442user185442
395
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
asked Mar 20 at 18:48
user185442user185442
395
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
edited Mar 20 at 18:55
Dr. Sonnhard Graubner
78.8k42867
78.8k42867
asked Mar 20 at 18:48
user185442user185442
395
asked Mar 20 at 18:48
user185442user185442
395
asked Mar 20 at 18:48
user185442user185442
395
395
$begingroup$
Is this right so?
$endgroup$
– Dr. Sonnhard Graubner
Mar 20 at 18:55
$begingroup$
yes sir, it is and answer is ((x tan x -1)/(x + tan x)) +(1/3)*(((x tan x -1)^3)/(x + tan x)) + c but i do not know the solution
$endgroup$
– user185442
Mar 21 at 17:42
add a comment |
$begingroup$
Is this right so?
$endgroup$
– Dr. Sonnhard Graubner
Mar 20 at 18:55
$begingroup$
yes sir, it is and answer is ((x tan x -1)/(x + tan x)) +(1/3)*(((x tan x -1)^3)/(x + tan x)) + c but i do not know the solution
$endgroup$
– user185442
Mar 21 at 17:42
$begingroup$
Is this right so?
$endgroup$
– Dr. Sonnhard Graubner
Mar 20 at 18:55
$begingroup$
Is this right so?
$endgroup$
– Dr. Sonnhard Graubner
Mar 20 at 18:55
$begingroup$
yes sir, it is and answer is ((x tan x -1)/(x + tan x)) +(1/3)*(((x tan x -1)^3)/(x + tan x)) + c but i do not know the solution
$endgroup$
– user185442
Mar 21 at 17:42
$begingroup$
yes sir, it is and answer is ((x tan x -1)/(x + tan x)) +(1/3)*(((x tan x -1)^3)/(x + tan x)) + c but i do not know the solution
$endgroup$
– user185442
Mar 21 at 17:42
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
put x = tan y ,then it will become (sec^4 y)(1+ sec^2 y)dx/((tan y)(cos tan y)+ sin tan y)
then take down sec^4 y in denominator and use sin(A+B) identity, the variable x in the sin(x) in denominator will have differential coefficient in numerator
answered Apr 5 at 16:35
user185442user185442
395
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
put x = tan y ,then it will become (sec^4 y)(1+ sec^2 y)dx/((tan y)(cos tan y)+ sin tan y)
then take down sec^4 y in denominator and use sin(A+B) identity, the variable x in the sin(x) in denominator will have differential coefficient in numerator
answered Apr 5 at 16:35
user185442user185442
395
$endgroup$
add a comment |
$begingroup$
put x = tan y ,then it will become (sec^4 y)(1+ sec^2 y)dx/((tan y)(cos tan y)+ sin tan y)
then take down sec^4 y in denominator and use sin(A+B) identity, the variable x in the sin(x) in denominator will have differential coefficient in numerator
answered Apr 5 at 16:35
user185442user185442
395
$endgroup$
add a comment |
$begingroup$
put x = tan y ,then it will become (sec^4 y)(1+ sec^2 y)dx/((tan y)(cos tan y)+ sin tan y)
then take down sec^4 y in denominator and use sin(A+B) identity, the variable x in the sin(x) in denominator will have differential coefficient in numerator
answered Apr 5 at 16:35
user185442user185442
395
$endgroup$
put x = tan y ,then it will become (sec^4 y)(1+ sec^2 y)dx/((tan y)(cos tan y)+ sin tan y)
then take down sec^4 y in denominator and use sin(A+B) identity, the variable x in the sin(x) in denominator will have differential coefficient in numerator
answered Apr 5 at 16:35
user185442user185442
395
answered Apr 5 at 16:35
user185442user185442
395
answered Apr 5 at 16:35
user185442user185442
395
answered Apr 5 at 16:35
user185442user185442
395
395
add a comment |
add a comment |
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$begingroup$
Is this right so?
$endgroup$
– Dr. Sonnhard Graubner
Mar 20 at 18:55
$begingroup$
yes sir, it is and answer is ((x tan x -1)/(x + tan x)) +(1/3)*(((x tan x -1)^3)/(x + tan x)) + c but i do not know the solution
$endgroup$
– user185442
Mar 21 at 17:42