Show that the substitution $t=tantheta$ transforms the integral ${int}frac{dtheta}{9cos^2theta+sin^2theta}$,...











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To begin with the $dtheta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:



$${int}frac{1}{9cos^2theta+sin^2theta},dtheta$$



I tried working backwards



$$frac{d}{dtheta}tantheta=sec^2theta,,,,{Rightarrow},,,,d,tantheta=sec^2theta,dtheta$$
$${Rightarrow},{int}frac{sec^2theta,dtheta}{9+tan^2theta}$$



$$tantheta=frac{sintheta}{costheta},,,,{Rightarrow},,,,tan^2theta=frac{sin^2theta}{cos^2theta}$$



$${Rightarrow},{int}frac{sec^2theta,dtheta}{left(9+dfrac{sin^2theta}{cos^2theta}right)}$$



$$9=frac{9cos^2theta}{cos^2theta},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta}{cos^2theta}+dfrac{sin^2theta}{cos^2theta}right)},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta+sin^2theta}{cos^2theta}right)}$$



$${Rightarrow},,,,{int}frac{color{red}{cos^2theta,sec^2theta},dtheta}{9cos^2theta+sin^2theta}$$



Now I have to prove that $$cos^2theta,sec^2theta=1$$
but I don't think it is... What have I done wrong? Regards Tom






trigonometry







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  • 4




    Define $costheta,sectheta$
    – lab bhattacharjee
    Apr 8 '15 at 11:49






  • 2




    $sec^2theta=frac{1}{cos^2theta} implies cos^2theta sec^2theta = 1$
    – Waffle
    Apr 8 '15 at 11:56










  • @Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue?
    – Robert Howard
    Nov 15 at 5:14










  • @Waffle Thanks!
    – Robert Howard
    Nov 16 at 1:00















up vote
3
down vote

favorite
2












To begin with the $dtheta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:



$${int}frac{1}{9cos^2theta+sin^2theta},dtheta$$



I tried working backwards



$$frac{d}{dtheta}tantheta=sec^2theta,,,,{Rightarrow},,,,d,tantheta=sec^2theta,dtheta$$
$${Rightarrow},{int}frac{sec^2theta,dtheta}{9+tan^2theta}$$



$$tantheta=frac{sintheta}{costheta},,,,{Rightarrow},,,,tan^2theta=frac{sin^2theta}{cos^2theta}$$



$${Rightarrow},{int}frac{sec^2theta,dtheta}{left(9+dfrac{sin^2theta}{cos^2theta}right)}$$



$$9=frac{9cos^2theta}{cos^2theta},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta}{cos^2theta}+dfrac{sin^2theta}{cos^2theta}right)},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta+sin^2theta}{cos^2theta}right)}$$



$${Rightarrow},,,,{int}frac{color{red}{cos^2theta,sec^2theta},dtheta}{9cos^2theta+sin^2theta}$$



Now I have to prove that $$cos^2theta,sec^2theta=1$$
but I don't think it is... What have I done wrong? Regards Tom






trigonometry







share|cite|improve this question




















  • 4




    Define $costheta,sectheta$
    – lab bhattacharjee
    Apr 8 '15 at 11:49






  • 2




    $sec^2theta=frac{1}{cos^2theta} implies cos^2theta sec^2theta = 1$
    – Waffle
    Apr 8 '15 at 11:56










  • @Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue?
    – Robert Howard
    Nov 15 at 5:14










  • @Waffle Thanks!
    – Robert Howard
    Nov 16 at 1:00













up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





To begin with the $dtheta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:



$${int}frac{1}{9cos^2theta+sin^2theta},dtheta$$



I tried working backwards



$$frac{d}{dtheta}tantheta=sec^2theta,,,,{Rightarrow},,,,d,tantheta=sec^2theta,dtheta$$
$${Rightarrow},{int}frac{sec^2theta,dtheta}{9+tan^2theta}$$



$$tantheta=frac{sintheta}{costheta},,,,{Rightarrow},,,,tan^2theta=frac{sin^2theta}{cos^2theta}$$



$${Rightarrow},{int}frac{sec^2theta,dtheta}{left(9+dfrac{sin^2theta}{cos^2theta}right)}$$



$$9=frac{9cos^2theta}{cos^2theta},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta}{cos^2theta}+dfrac{sin^2theta}{cos^2theta}right)},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta+sin^2theta}{cos^2theta}right)}$$



$${Rightarrow},,,,{int}frac{color{red}{cos^2theta,sec^2theta},dtheta}{9cos^2theta+sin^2theta}$$



Now I have to prove that $$cos^2theta,sec^2theta=1$$
but I don't think it is... What have I done wrong? Regards Tom






trigonometry







share|cite|improve this question















To begin with the $dtheta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:



$${int}frac{1}{9cos^2theta+sin^2theta},dtheta$$



I tried working backwards



$$frac{d}{dtheta}tantheta=sec^2theta,,,,{Rightarrow},,,,d,tantheta=sec^2theta,dtheta$$
$${Rightarrow},{int}frac{sec^2theta,dtheta}{9+tan^2theta}$$



$$tantheta=frac{sintheta}{costheta},,,,{Rightarrow},,,,tan^2theta=frac{sin^2theta}{cos^2theta}$$



$${Rightarrow},{int}frac{sec^2theta,dtheta}{left(9+dfrac{sin^2theta}{cos^2theta}right)}$$



$$9=frac{9cos^2theta}{cos^2theta},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta}{cos^2theta}+dfrac{sin^2theta}{cos^2theta}right)},,,,{Rightarrow},,,,{int}frac{sec^2theta,dtheta}{left(dfrac{9cos^2theta+sin^2theta}{cos^2theta}right)}$$



$${Rightarrow},,,,{int}frac{color{red}{cos^2theta,sec^2theta},dtheta}{9cos^2theta+sin^2theta}$$



Now I have to prove that $$cos^2theta,sec^2theta=1$$
but I don't think it is... What have I done wrong? Regards Tom






trigonometry




trigonometry






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edited Apr 8 '15 at 13:06









N. F. Taussig

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asked Apr 8 '15 at 11:47









Thomas Winkworth

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  • 4




    Define $costheta,sectheta$
    – lab bhattacharjee
    Apr 8 '15 at 11:49






  • 2




    $sec^2theta=frac{1}{cos^2theta} implies cos^2theta sec^2theta = 1$
    – Waffle
    Apr 8 '15 at 11:56










  • @Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue?
    – Robert Howard
    Nov 15 at 5:14










  • @Waffle Thanks!
    – Robert Howard
    Nov 16 at 1:00














  • 4




    Define $costheta,sectheta$
    – lab bhattacharjee
    Apr 8 '15 at 11:49






  • 2




    $sec^2theta=frac{1}{cos^2theta} implies cos^2theta sec^2theta = 1$
    – Waffle
    Apr 8 '15 at 11:56










  • @Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue?
    – Robert Howard
    Nov 15 at 5:14










  • @Waffle Thanks!
    – Robert Howard
    Nov 16 at 1:00








4




4




Define $costheta,sectheta$
– lab bhattacharjee
Apr 8 '15 at 11:49




Define $costheta,sectheta$
– lab bhattacharjee
Apr 8 '15 at 11:49




2




2




$sec^2theta=frac{1}{cos^2theta} implies cos^2theta sec^2theta = 1$
– Waffle
Apr 8 '15 at 11:56




$sec^2theta=frac{1}{cos^2theta} implies cos^2theta sec^2theta = 1$
– Waffle
Apr 8 '15 at 11:56












@Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue?
– Robert Howard
Nov 15 at 5:14




@Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue?
– Robert Howard
Nov 15 at 5:14












@Waffle Thanks!
– Robert Howard
Nov 16 at 1:00




@Waffle Thanks!
– Robert Howard
Nov 16 at 1:00










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$sec^2theta = frac{1}{cos^2theta} implies cos^2thetasec^2theta = frac{cos^2theta}{cos^2theta} = 1$






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    $sec^2theta = frac{1}{cos^2theta} implies cos^2thetasec^2theta = frac{cos^2theta}{cos^2theta} = 1$






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      $sec^2theta = frac{1}{cos^2theta} implies cos^2thetasec^2theta = frac{cos^2theta}{cos^2theta} = 1$






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        $sec^2theta = frac{1}{cos^2theta} implies cos^2thetasec^2theta = frac{cos^2theta}{cos^2theta} = 1$






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        $sec^2theta = frac{1}{cos^2theta} implies cos^2thetasec^2theta = frac{cos^2theta}{cos^2theta} = 1$







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        answered Nov 15 at 19:55









        Waffle

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