Integrate :- $int dx/(sin(x) + asec(x))^2$

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3
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Guys please help me in evaluating this integral
$$
int frac1(sin(x) + a sec(x))^2,dx
$$
I tried by converting $sec(x)$ to $cos(x)$ and by solving it became more complicated so guys please guide me further.




calculus integration indefinite-integrals




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  • 1




    I have no idea how but take a look at it : [wolframalpha.com/input/…
    – mrtaurho
    Aug 3 at 11:21











  • Have you tried Weierstrauss sub?
    – Henry Lee
    15 hours ago














up vote
3
down vote

favorite












Guys please help me in evaluating this integral
$$
int frac1(sin(x) + a sec(x))^2,dx
$$
I tried by converting $sec(x)$ to $cos(x)$ and by solving it became more complicated so guys please guide me further.




calculus integration indefinite-integrals




share|cite|improve this question

















  • 1




    I have no idea how but take a look at it : [wolframalpha.com/input/…
    – mrtaurho
    Aug 3 at 11:21











  • Have you tried Weierstrauss sub?
    – Henry Lee
    15 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Guys please help me in evaluating this integral
$$
int frac1(sin(x) + a sec(x))^2,dx
$$
I tried by converting $sec(x)$ to $cos(x)$ and by solving it became more complicated so guys please guide me further.




calculus integration indefinite-integrals




share|cite|improve this question













Guys please help me in evaluating this integral
$$
int frac1(sin(x) + a sec(x))^2,dx
$$
I tried by converting $sec(x)$ to $cos(x)$ and by solving it became more complicated so guys please guide me further.





calculus integration indefinite-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 18:44









Abcd

2,3101524




2,3101524









asked Aug 3 at 11:13









Ritik

4310




4310







  • 1




    I have no idea how but take a look at it : [wolframalpha.com/input/…
    – mrtaurho
    Aug 3 at 11:21











  • Have you tried Weierstrauss sub?
    – Henry Lee
    15 hours ago












  • 1




    I have no idea how but take a look at it : [wolframalpha.com/input/…
    – mrtaurho
    Aug 3 at 11:21











  • Have you tried Weierstrauss sub?
    – Henry Lee
    15 hours ago







1




1




I have no idea how but take a look at it : [wolframalpha.com/input/…
– mrtaurho
Aug 3 at 11:21





I have no idea how but take a look at it : [wolframalpha.com/input/…
– mrtaurho
Aug 3 at 11:21













Have you tried Weierstrauss sub?
– Henry Lee
15 hours ago




Have you tried Weierstrauss sub?
– Henry Lee
15 hours ago










1 Answer
1






active

oldest

votes

















up vote
6
down vote



accepted










Hint:
$$dfrac1(sin x+asec x)^2=dfrac12(sin xcos x+a)^2+dfraccos2x2(sin xcos x+a)^2$$



The second part is elementary.



$$dfrac1(sin xcos x+a)^2=dfracsec^2x(1+tan^2x)(tan x+atan^2x+a)^2$$



Choose $tan x=u$






share|cite|improve this answer





















  • I didn't get what You did after saying elementary
    – Ritik
    Aug 3 at 11:35






  • 2




    @Ritik, Divided numerator & denominator by $$cos^4x$$
    – lab bhattacharjee
    Aug 3 at 11:37






  • 1




    For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
    – egreg
    Aug 3 at 11:46







  • 1




    How you guys are so good at integration ??
    – Ritik
    Aug 3 at 11:50






  • 1




    @labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
    – egreg
    Aug 3 at 13:47










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote



accepted










Hint:
$$dfrac1(sin x+asec x)^2=dfrac12(sin xcos x+a)^2+dfraccos2x2(sin xcos x+a)^2$$



The second part is elementary.



$$dfrac1(sin xcos x+a)^2=dfracsec^2x(1+tan^2x)(tan x+atan^2x+a)^2$$



Choose $tan x=u$






share|cite|improve this answer





















  • I didn't get what You did after saying elementary
    – Ritik
    Aug 3 at 11:35






  • 2




    @Ritik, Divided numerator & denominator by $$cos^4x$$
    – lab bhattacharjee
    Aug 3 at 11:37






  • 1




    For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
    – egreg
    Aug 3 at 11:46







  • 1




    How you guys are so good at integration ??
    – Ritik
    Aug 3 at 11:50






  • 1




    @labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
    – egreg
    Aug 3 at 13:47














up vote
6
down vote



accepted










Hint:
$$dfrac1(sin x+asec x)^2=dfrac12(sin xcos x+a)^2+dfraccos2x2(sin xcos x+a)^2$$



The second part is elementary.



$$dfrac1(sin xcos x+a)^2=dfracsec^2x(1+tan^2x)(tan x+atan^2x+a)^2$$



Choose $tan x=u$






share|cite|improve this answer





















  • I didn't get what You did after saying elementary
    – Ritik
    Aug 3 at 11:35






  • 2




    @Ritik, Divided numerator & denominator by $$cos^4x$$
    – lab bhattacharjee
    Aug 3 at 11:37






  • 1




    For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
    – egreg
    Aug 3 at 11:46







  • 1




    How you guys are so good at integration ??
    – Ritik
    Aug 3 at 11:50






  • 1




    @labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
    – egreg
    Aug 3 at 13:47












up vote
6
down vote



accepted







up vote
6
down vote



accepted






Hint:
$$dfrac1(sin x+asec x)^2=dfrac12(sin xcos x+a)^2+dfraccos2x2(sin xcos x+a)^2$$



The second part is elementary.



$$dfrac1(sin xcos x+a)^2=dfracsec^2x(1+tan^2x)(tan x+atan^2x+a)^2$$



Choose $tan x=u$






share|cite|improve this answer













Hint:
$$dfrac1(sin x+asec x)^2=dfrac12(sin xcos x+a)^2+dfraccos2x2(sin xcos x+a)^2$$



The second part is elementary.



$$dfrac1(sin xcos x+a)^2=dfracsec^2x(1+tan^2x)(tan x+atan^2x+a)^2$$



Choose $tan x=u$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 11:24









lab bhattacharjee

214k14152263




214k14152263











  • I didn't get what You did after saying elementary
    – Ritik
    Aug 3 at 11:35






  • 2




    @Ritik, Divided numerator & denominator by $$cos^4x$$
    – lab bhattacharjee
    Aug 3 at 11:37






  • 1




    For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
    – egreg
    Aug 3 at 11:46







  • 1




    How you guys are so good at integration ??
    – Ritik
    Aug 3 at 11:50






  • 1




    @labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
    – egreg
    Aug 3 at 13:47
















  • I didn't get what You did after saying elementary
    – Ritik
    Aug 3 at 11:35






  • 2




    @Ritik, Divided numerator & denominator by $$cos^4x$$
    – lab bhattacharjee
    Aug 3 at 11:37






  • 1




    For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
    – egreg
    Aug 3 at 11:46







  • 1




    How you guys are so good at integration ??
    – Ritik
    Aug 3 at 11:50






  • 1




    @labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
    – egreg
    Aug 3 at 13:47















I didn't get what You did after saying elementary
– Ritik
Aug 3 at 11:35




I didn't get what You did after saying elementary
– Ritik
Aug 3 at 11:35




2




2




@Ritik, Divided numerator & denominator by $$cos^4x$$
– lab bhattacharjee
Aug 3 at 11:37




@Ritik, Divided numerator & denominator by $$cos^4x$$
– lab bhattacharjee
Aug 3 at 11:37




1




1




For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
– egreg
Aug 3 at 11:46





For the second fraction, it is simpler to set $u=sin2x$, so $du=2cos2x$ and the integral becomes $intfrac1(u/2+a)^2,du$
– egreg
Aug 3 at 11:46





1




1




How you guys are so good at integration ??
– Ritik
Aug 3 at 11:50




How you guys are so good at integration ??
– Ritik
Aug 3 at 11:50




1




1




@labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
– egreg
Aug 3 at 13:47




@labbhattacharjee I don't think that dividing by $cos^4x$ is the best way.
– egreg
Aug 3 at 13:47












 

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