How to integrate this indefinite integral?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite
1












$$intfracx^2sec^2x(x tanx+1)^2,mathrmdx$$



I tried the online available calculators but they cannot calculate the answer or provide the solution.




calculus indefinite-integrals




share|cite|improve this question

















  • 1




    What online calculators are you using, exactly? WolframAlpha does it smoothly.
    – Arthur
    Aug 3 at 18:03











  • Wolfram cannot calculate the solution though :(
    – Chirag Mahajan
    Aug 3 at 18:04






  • 2




    You mean show you the steps? Technically it can (I think it costs money), but you don't need to. Take the answer as given on WA, differentiate it, get $fracx^2sec^2x(xtan x+1)^2$, then do the steps in reverse, and you have your solution.
    – Arthur
    Aug 3 at 18:06







  • 2




    @bobcliffe your edit is not useful.
    – user 108128
    Aug 3 at 18:21














up vote
3
down vote

favorite
1












$$intfracx^2sec^2x(x tanx+1)^2,mathrmdx$$



I tried the online available calculators but they cannot calculate the answer or provide the solution.




calculus indefinite-integrals




share|cite|improve this question

















  • 1




    What online calculators are you using, exactly? WolframAlpha does it smoothly.
    – Arthur
    Aug 3 at 18:03











  • Wolfram cannot calculate the solution though :(
    – Chirag Mahajan
    Aug 3 at 18:04






  • 2




    You mean show you the steps? Technically it can (I think it costs money), but you don't need to. Take the answer as given on WA, differentiate it, get $fracx^2sec^2x(xtan x+1)^2$, then do the steps in reverse, and you have your solution.
    – Arthur
    Aug 3 at 18:06







  • 2




    @bobcliffe your edit is not useful.
    – user 108128
    Aug 3 at 18:21












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





$$intfracx^2sec^2x(x tanx+1)^2,mathrmdx$$



I tried the online available calculators but they cannot calculate the answer or provide the solution.




calculus indefinite-integrals




share|cite|improve this question













$$intfracx^2sec^2x(x tanx+1)^2,mathrmdx$$



I tried the online available calculators but they cannot calculate the answer or provide the solution.





calculus indefinite-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 18:28









Robert Howard

1,241620




1,241620









asked Aug 3 at 18:01









Chirag Mahajan

214




214







  • 1




    What online calculators are you using, exactly? WolframAlpha does it smoothly.
    – Arthur
    Aug 3 at 18:03











  • Wolfram cannot calculate the solution though :(
    – Chirag Mahajan
    Aug 3 at 18:04






  • 2




    You mean show you the steps? Technically it can (I think it costs money), but you don't need to. Take the answer as given on WA, differentiate it, get $fracx^2sec^2x(xtan x+1)^2$, then do the steps in reverse, and you have your solution.
    – Arthur
    Aug 3 at 18:06







  • 2




    @bobcliffe your edit is not useful.
    – user 108128
    Aug 3 at 18:21












  • 1




    What online calculators are you using, exactly? WolframAlpha does it smoothly.
    – Arthur
    Aug 3 at 18:03











  • Wolfram cannot calculate the solution though :(
    – Chirag Mahajan
    Aug 3 at 18:04






  • 2




    You mean show you the steps? Technically it can (I think it costs money), but you don't need to. Take the answer as given on WA, differentiate it, get $fracx^2sec^2x(xtan x+1)^2$, then do the steps in reverse, and you have your solution.
    – Arthur
    Aug 3 at 18:06







  • 2




    @bobcliffe your edit is not useful.
    – user 108128
    Aug 3 at 18:21







1




1




What online calculators are you using, exactly? WolframAlpha does it smoothly.
– Arthur
Aug 3 at 18:03





What online calculators are you using, exactly? WolframAlpha does it smoothly.
– Arthur
Aug 3 at 18:03













Wolfram cannot calculate the solution though :(
– Chirag Mahajan
Aug 3 at 18:04




Wolfram cannot calculate the solution though :(
– Chirag Mahajan
Aug 3 at 18:04




2




2




You mean show you the steps? Technically it can (I think it costs money), but you don't need to. Take the answer as given on WA, differentiate it, get $fracx^2sec^2x(xtan x+1)^2$, then do the steps in reverse, and you have your solution.
– Arthur
Aug 3 at 18:06





You mean show you the steps? Technically it can (I think it costs money), but you don't need to. Take the answer as given on WA, differentiate it, get $fracx^2sec^2x(xtan x+1)^2$, then do the steps in reverse, and you have your solution.
– Arthur
Aug 3 at 18:06





2




2




@bobcliffe your edit is not useful.
– user 108128
Aug 3 at 18:21




@bobcliffe your edit is not useful.
– user 108128
Aug 3 at 18:21










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










This is basically working backwards, like @Arthur suggests:



beginalign
fracx^2sec^2 x(xtan x+1)^2 &= fracx^2left(xfracsin xcos x + 1right)^2cos^2 x\
&= fracx^2(xsin x + cos x)^2\
&= fracx^2(sin^2 x + cos ^2 x) - xcos xsin x + x cos xsin x(xsin x + cos x)^2\
&= fracxsin x(x sin x + cos x) - (sin x - xcos x)xcos x(xsin x + cos x)^2\
&= frac(sin x - xcos x)'(x sin x + cos x) - (sin x - xcos x)(xsin x + cos x)'(xsin x + cos x)^2\
&= left(fracsin x - xcos xx sin x + cos xright)'
endalign



so $$int fracx^2sec^2 x(xtan x+1)^2 ,dx = fracsin x - xcos xx sin x + cos x$$






share|cite|improve this answer

















  • 4




    Don't you just love it when the numerator takes exactly as much space as the denominator? :)
    – TheSimpliFire
    Aug 3 at 19:22










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871344%2fhow-to-integrate-this-indefinite-integral%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










This is basically working backwards, like @Arthur suggests:



beginalign
fracx^2sec^2 x(xtan x+1)^2 &= fracx^2left(xfracsin xcos x + 1right)^2cos^2 x\
&= fracx^2(xsin x + cos x)^2\
&= fracx^2(sin^2 x + cos ^2 x) - xcos xsin x + x cos xsin x(xsin x + cos x)^2\
&= fracxsin x(x sin x + cos x) - (sin x - xcos x)xcos x(xsin x + cos x)^2\
&= frac(sin x - xcos x)'(x sin x + cos x) - (sin x - xcos x)(xsin x + cos x)'(xsin x + cos x)^2\
&= left(fracsin x - xcos xx sin x + cos xright)'
endalign



so $$int fracx^2sec^2 x(xtan x+1)^2 ,dx = fracsin x - xcos xx sin x + cos x$$






share|cite|improve this answer

















  • 4




    Don't you just love it when the numerator takes exactly as much space as the denominator? :)
    – TheSimpliFire
    Aug 3 at 19:22














up vote
2
down vote



accepted










This is basically working backwards, like @Arthur suggests:



beginalign
fracx^2sec^2 x(xtan x+1)^2 &= fracx^2left(xfracsin xcos x + 1right)^2cos^2 x\
&= fracx^2(xsin x + cos x)^2\
&= fracx^2(sin^2 x + cos ^2 x) - xcos xsin x + x cos xsin x(xsin x + cos x)^2\
&= fracxsin x(x sin x + cos x) - (sin x - xcos x)xcos x(xsin x + cos x)^2\
&= frac(sin x - xcos x)'(x sin x + cos x) - (sin x - xcos x)(xsin x + cos x)'(xsin x + cos x)^2\
&= left(fracsin x - xcos xx sin x + cos xright)'
endalign



so $$int fracx^2sec^2 x(xtan x+1)^2 ,dx = fracsin x - xcos xx sin x + cos x$$






share|cite|improve this answer

















  • 4




    Don't you just love it when the numerator takes exactly as much space as the denominator? :)
    – TheSimpliFire
    Aug 3 at 19:22












up vote
2
down vote



accepted







up vote
2
down vote



accepted






This is basically working backwards, like @Arthur suggests:



beginalign
fracx^2sec^2 x(xtan x+1)^2 &= fracx^2left(xfracsin xcos x + 1right)^2cos^2 x\
&= fracx^2(xsin x + cos x)^2\
&= fracx^2(sin^2 x + cos ^2 x) - xcos xsin x + x cos xsin x(xsin x + cos x)^2\
&= fracxsin x(x sin x + cos x) - (sin x - xcos x)xcos x(xsin x + cos x)^2\
&= frac(sin x - xcos x)'(x sin x + cos x) - (sin x - xcos x)(xsin x + cos x)'(xsin x + cos x)^2\
&= left(fracsin x - xcos xx sin x + cos xright)'
endalign



so $$int fracx^2sec^2 x(xtan x+1)^2 ,dx = fracsin x - xcos xx sin x + cos x$$






share|cite|improve this answer













This is basically working backwards, like @Arthur suggests:



beginalign
fracx^2sec^2 x(xtan x+1)^2 &= fracx^2left(xfracsin xcos x + 1right)^2cos^2 x\
&= fracx^2(xsin x + cos x)^2\
&= fracx^2(sin^2 x + cos ^2 x) - xcos xsin x + x cos xsin x(xsin x + cos x)^2\
&= fracxsin x(x sin x + cos x) - (sin x - xcos x)xcos x(xsin x + cos x)^2\
&= frac(sin x - xcos x)'(x sin x + cos x) - (sin x - xcos x)(xsin x + cos x)'(xsin x + cos x)^2\
&= left(fracsin x - xcos xx sin x + cos xright)'
endalign



so $$int fracx^2sec^2 x(xtan x+1)^2 ,dx = fracsin x - xcos xx sin x + cos x$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 19:17









mechanodroid

22.2k52041




22.2k52041







  • 4




    Don't you just love it when the numerator takes exactly as much space as the denominator? :)
    – TheSimpliFire
    Aug 3 at 19:22












  • 4




    Don't you just love it when the numerator takes exactly as much space as the denominator? :)
    – TheSimpliFire
    Aug 3 at 19:22







4




4




Don't you just love it when the numerator takes exactly as much space as the denominator? :)
– TheSimpliFire
Aug 3 at 19:22




Don't you just love it when the numerator takes exactly as much space as the denominator? :)
– TheSimpliFire
Aug 3 at 19:22












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871344%2fhow-to-integrate-this-indefinite-integral%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more