indefinite integration of polynomial and trignometric functionsindefinite integral of $x^nsin(x)$Integrate $int_0^pixsin xover1+cos^2xdx$.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 $int e^2cos xcos 2x dx$How to solve this integration problem by parts and substitution?Indefinite Integration of a trig functionIntegral of $int_0^b cos(x)cos(frac a x)dx$

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indefinite integration of polynomial and trignometric functions


indefinite integral of $x^nsin(x)$Integrate $int_0^pixsin xover1+cos^2xdx$.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 $int e^2cos xcos 2x dx$How to solve this integration problem by parts and substitution?Indefinite Integration of a trig functionIntegral of $int_0^b cos(x)cos(frac a x)dx$













1












$begingroup$


integrate $$int frac(1+ x^2)(2+ x^2)(x cos x + sin x)^4dx$$
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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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















1












$begingroup$


integrate $$int frac(1+ x^2)(2+ x^2)(x cos x + sin x)^4dx$$
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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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













1












1








1


1



$begingroup$


integrate $$int frac(1+ x^2)(2+ x^2)(x cos x + sin x)^4dx$$
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.










share|cite|improve this question











$endgroup$




integrate $$int frac(1+ x^2)(2+ x^2)(x cos x + sin x)^4dx$$
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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 18:55









Dr. Sonnhard Graubner

78.4k42867




78.4k42867










asked Mar 20 at 18:48









user185442user185442

192




192











  • $begingroup$
    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










  • $begingroup$
    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










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