An integral that needs subtitution to be solved.Integral $int tan^5(x)text dx$How to evaluate this indefinite integration $int fractan^4 theta d theta1-tan^2 theta$?Integral $int fracdxtan x + cot x + csc x + sec x$$int(sec^4x)dx = cdots$Evaluating $intfraccos^2x1+tan x,dx$Alternative Integral solution to $sec^2(x) sin(x)$Integration changing limits - does the question have an error?Example integral question that contains more than 2 Integral TechniquesCan these integrals be solved by u-substitution using trig identities?Find the integral $int_sec^3x$ via u-substitution
Was it ever illegal to name a pig "Napoleon" in France?
What does the multimeter dial do internally?
Are all diatonic chords in the diminished scale diminished?
No Torah = Revert to Nothingness?
Conditions for Roots of a quadratic equation at infinity
What does collachrimation mean?
Intern not wearing safety equipment; how could I have handled this differently?
VHF 50 Ω Antenna Over 75 Ω TV Coax
How should I ask for a "pint" in countries that use metric?
What are the effects of abstaining from eating a certain flavor?
Is it better in terms of durability to remove card+battery or to connect to charger/computer via USB-C?
What exactly is a "murder hobo"?
Why does the Antonov AN-225 not have any winglets?
Is there a method for differentiating informative comments from commented out code?
Party going through airport security at separate times?
Would a carnivorous diet be able to support a giant worm?
What is a writing material that persists nearly forever or for a long time?
Need a non-volatile memory IC with near unlimited read/write operations capability
Password Hashing Security Using Scrypt & Argon2
When I press the space bar it deletes the letters in front of it
How many tone holes are there actually in different orchestral woodwind instruments?
Did the Ottoman empire suppress the printing press?
Is it ok for parents to kiss and romance with each other while their 2- to 8-year-old child watches?
Is it okay to roll multiple attacks that all have advantage in one cluster?
An integral that needs subtitution to be solved.
Integral $int tan^5(x)text dx$How to evaluate this indefinite integration $int fractan^4 theta d theta1-tan^2 theta$?Integral $int fracdxtan x + cot x + csc x + sec x$$int(sec^4x)dx = cdots$Evaluating $intfraccos^2x1+tan x,dx$Alternative Integral solution to $sec^2(x) sin(x)$Integration changing limits - does the question have an error?Example integral question that contains more than 2 Integral TechniquesCan these integrals be solved by u-substitution using trig identities?Find the integral $int_sec^3x$ via u-substitution
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I have this excercise. The hint is that it's solved by the substitution method, but I'm not able to find the right one, or even get close to something.
$$intfracsec(x)sin(x)+cos(x)dx$$
Using identities, I have arrived to things like this:
$$-intfractan(x)-1cos(2x)dx$$
But, well, the thing above doesn't help to much.
I'd appreciate any help on this subject.
Good day to the people that read this.
integration trigonometric-integrals
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
$endgroup$
add a comment |
$begingroup$
I have this excercise. The hint is that it's solved by the substitution method, but I'm not able to find the right one, or even get close to something.
$$intfracsec(x)sin(x)+cos(x)dx$$
Using identities, I have arrived to things like this:
$$-intfractan(x)-1cos(2x)dx$$
But, well, the thing above doesn't help to much.
I'd appreciate any help on this subject.
Good day to the people that read this.
integration trigonometric-integrals
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
$endgroup$
$begingroup$
There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2pi/n$, substitute $t=tanfracnx2$. In this case $n=2$, so $sec xdx=cos xdt$. The rest writes itself.
$endgroup$
– J.G.
8 hours ago
add a comment |
$begingroup$
I have this excercise. The hint is that it's solved by the substitution method, but I'm not able to find the right one, or even get close to something.
$$intfracsec(x)sin(x)+cos(x)dx$$
Using identities, I have arrived to things like this:
$$-intfractan(x)-1cos(2x)dx$$
But, well, the thing above doesn't help to much.
I'd appreciate any help on this subject.
Good day to the people that read this.
integration trigonometric-integrals
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
$endgroup$
I have this excercise. The hint is that it's solved by the substitution method, but I'm not able to find the right one, or even get close to something.
$$intfracsec(x)sin(x)+cos(x)dx$$
Using identities, I have arrived to things like this:
$$-intfractan(x)-1cos(2x)dx$$
But, well, the thing above doesn't help to much.
I'd appreciate any help on this subject.
Good day to the people that read this.
integration trigonometric-integrals
integration trigonometric-integrals
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
edited 9 hours ago
José Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
198k24 gold badges156 silver badges273 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
asked 9 hours ago
ScoofjeerScoofjeer
1405 bronze badges
1405 bronze badges
$begingroup$
There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2pi/n$, substitute $t=tanfracnx2$. In this case $n=2$, so $sec xdx=cos xdt$. The rest writes itself.
$endgroup$
– J.G.
8 hours ago
add a comment |
$begingroup$
There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2pi/n$, substitute $t=tanfracnx2$. In this case $n=2$, so $sec xdx=cos xdt$. The rest writes itself.
$endgroup$
– J.G.
8 hours ago
$begingroup$
There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2pi/n$, substitute $t=tanfracnx2$. In this case $n=2$, so $sec xdx=cos xdt$. The rest writes itself.
$endgroup$
– J.G.
8 hours ago
$begingroup$
There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2pi/n$, substitute $t=tanfracnx2$. In this case $n=2$, so $sec xdx=cos xdt$. The rest writes itself.
$endgroup$
– J.G.
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here's a solution using a substitution.
In particular, observe that $$fracsec xsin x+cos xcdotfracsec xsec x=fracsec^2 xtan x+1.$$ This is useful because $fracddxtan x=sec^2x$, so substituting $u=tan x$ tells us that $$intfracsec xsin x+cos xdx=intfrac1u+1du=ln(u+1)=ln(tan x+1).$$
Since $tan x=fracsin xcos x$, we find that this is simply equal to $$lnleft(fracsin x+cos xcos xright)=lnleft(|sin x+cos x|right)-lnleft(|cos x|right).$$
answered 9 hours ago
hiabchiabc
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
I hope that you don't mind if I don't use a substitution. Note thatbeginalignintfracsec(x)sin(x)+cos(x),mathrm dx&=intfrac1sin(x)cos(x)+cos^2(x),mathrm dx\&=intfraccos(x)-sin(x)cos(x)+sin(x)+fracsin(x)cos(x),mathrm dx\&=logbigl(lvertcos(x)+sin(x)rvertbigr)-logbigl(lvertcos(x)rvertbigr).endalign
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
$endgroup$
add a comment |
$begingroup$
You can observe that
$$
fracsec xsin x+cos x=frac1cos x(sin x+cos x)=
fracsin^2x+cos^2xcos x(sin x+cos x)=
fractan^2x+1tan x+1
$$
A rational function in the tangent can be integrated via
$$
u=tan x
$$
so $du=(1+tan^2x),dx$ or
$$
dx=frac1u^2+1,du
$$
and the integral is now elementary.
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
$endgroup$
add a comment |
Your Answer
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3286042%2fan-integral-that-needs-subtitution-to-be-solved%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a solution using a substitution.
In particular, observe that $$fracsec xsin x+cos xcdotfracsec xsec x=fracsec^2 xtan x+1.$$ This is useful because $fracddxtan x=sec^2x$, so substituting $u=tan x$ tells us that $$intfracsec xsin x+cos xdx=intfrac1u+1du=ln(u+1)=ln(tan x+1).$$
Since $tan x=fracsin xcos x$, we find that this is simply equal to $$lnleft(fracsin x+cos xcos xright)=lnleft(|sin x+cos x|right)-lnleft(|cos x|right).$$
answered 9 hours ago
hiabchiabc
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
Here's a solution using a substitution.
In particular, observe that $$fracsec xsin x+cos xcdotfracsec xsec x=fracsec^2 xtan x+1.$$ This is useful because $fracddxtan x=sec^2x$, so substituting $u=tan x$ tells us that $$intfracsec xsin x+cos xdx=intfrac1u+1du=ln(u+1)=ln(tan x+1).$$
Since $tan x=fracsin xcos x$, we find that this is simply equal to $$lnleft(fracsin x+cos xcos xright)=lnleft(|sin x+cos x|right)-lnleft(|cos x|right).$$
answered 9 hours ago
hiabchiabc
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
Here's a solution using a substitution.
In particular, observe that $$fracsec xsin x+cos xcdotfracsec xsec x=fracsec^2 xtan x+1.$$ This is useful because $fracddxtan x=sec^2x$, so substituting $u=tan x$ tells us that $$intfracsec xsin x+cos xdx=intfrac1u+1du=ln(u+1)=ln(tan x+1).$$
Since $tan x=fracsin xcos x$, we find that this is simply equal to $$lnleft(fracsin x+cos xcos xright)=lnleft(|sin x+cos x|right)-lnleft(|cos x|right).$$
answered 9 hours ago
hiabchiabc
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Here's a solution using a substitution.
In particular, observe that $$fracsec xsin x+cos xcdotfracsec xsec x=fracsec^2 xtan x+1.$$ This is useful because $fracddxtan x=sec^2x$, so substituting $u=tan x$ tells us that $$intfracsec xsin x+cos xdx=intfrac1u+1du=ln(u+1)=ln(tan x+1).$$
Since $tan x=fracsin xcos x$, we find that this is simply equal to $$lnleft(fracsin x+cos xcos xright)=lnleft(|sin x+cos x|right)-lnleft(|cos x|right).$$
answered 9 hours ago
hiabchiabc
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
answered 9 hours ago
hiabchiabc
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 9 hours ago
hiabchiabc
1016 bronze badges
answered 9 hours ago
hiabchiabc
1016 bronze badges
1016 bronze badges
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hiabc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
1
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
1
1
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
(+1) Nice! Your approach is better than mine.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
I hope that you don't mind if I don't use a substitution. Note thatbeginalignintfracsec(x)sin(x)+cos(x),mathrm dx&=intfrac1sin(x)cos(x)+cos^2(x),mathrm dx\&=intfraccos(x)-sin(x)cos(x)+sin(x)+fracsin(x)cos(x),mathrm dx\&=logbigl(lvertcos(x)+sin(x)rvertbigr)-logbigl(lvertcos(x)rvertbigr).endalign
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
$endgroup$
add a comment |
$begingroup$
I hope that you don't mind if I don't use a substitution. Note thatbeginalignintfracsec(x)sin(x)+cos(x),mathrm dx&=intfrac1sin(x)cos(x)+cos^2(x),mathrm dx\&=intfraccos(x)-sin(x)cos(x)+sin(x)+fracsin(x)cos(x),mathrm dx\&=logbigl(lvertcos(x)+sin(x)rvertbigr)-logbigl(lvertcos(x)rvertbigr).endalign
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
$endgroup$
add a comment |
$begingroup$
I hope that you don't mind if I don't use a substitution. Note thatbeginalignintfracsec(x)sin(x)+cos(x),mathrm dx&=intfrac1sin(x)cos(x)+cos^2(x),mathrm dx\&=intfraccos(x)-sin(x)cos(x)+sin(x)+fracsin(x)cos(x),mathrm dx\&=logbigl(lvertcos(x)+sin(x)rvertbigr)-logbigl(lvertcos(x)rvertbigr).endalign
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
$endgroup$
I hope that you don't mind if I don't use a substitution. Note thatbeginalignintfracsec(x)sin(x)+cos(x),mathrm dx&=intfrac1sin(x)cos(x)+cos^2(x),mathrm dx\&=intfraccos(x)-sin(x)cos(x)+sin(x)+fracsin(x)cos(x),mathrm dx\&=logbigl(lvertcos(x)+sin(x)rvertbigr)-logbigl(lvertcos(x)rvertbigr).endalign
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
198k24 gold badges156 silver badges273 bronze badges
198k24 gold badges156 silver badges273 bronze badges
add a comment |
add a comment |
$begingroup$
You can observe that
$$
fracsec xsin x+cos x=frac1cos x(sin x+cos x)=
fracsin^2x+cos^2xcos x(sin x+cos x)=
fractan^2x+1tan x+1
$$
A rational function in the tangent can be integrated via
$$
u=tan x
$$
so $du=(1+tan^2x),dx$ or
$$
dx=frac1u^2+1,du
$$
and the integral is now elementary.
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
$endgroup$
add a comment |
$begingroup$
You can observe that
$$
fracsec xsin x+cos x=frac1cos x(sin x+cos x)=
fracsin^2x+cos^2xcos x(sin x+cos x)=
fractan^2x+1tan x+1
$$
A rational function in the tangent can be integrated via
$$
u=tan x
$$
so $du=(1+tan^2x),dx$ or
$$
dx=frac1u^2+1,du
$$
and the integral is now elementary.
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
$endgroup$
add a comment |
$begingroup$
You can observe that
$$
fracsec xsin x+cos x=frac1cos x(sin x+cos x)=
fracsin^2x+cos^2xcos x(sin x+cos x)=
fractan^2x+1tan x+1
$$
A rational function in the tangent can be integrated via
$$
u=tan x
$$
so $du=(1+tan^2x),dx$ or
$$
dx=frac1u^2+1,du
$$
and the integral is now elementary.
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
$endgroup$
You can observe that
$$
fracsec xsin x+cos x=frac1cos x(sin x+cos x)=
fracsin^2x+cos^2xcos x(sin x+cos x)=
fractan^2x+1tan x+1
$$
A rational function in the tangent can be integrated via
$$
u=tan x
$$
so $du=(1+tan^2x),dx$ or
$$
dx=frac1u^2+1,du
$$
and the integral is now elementary.
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
answered 9 hours ago
egregegreg
190k14 gold badges90 silver badges213 bronze badges
190k14 gold badges90 silver badges213 bronze badges
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3286042%2fan-integral-that-needs-subtitution-to-be-solved%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2pi/n$, substitute $t=tanfracnx2$. In this case $n=2$, so $sec xdx=cos xdt$. The rest writes itself.
$endgroup$
– J.G.
8 hours ago