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

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

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

add a comment |

up vote
3
down vote

favorite

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

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

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

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

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

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

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

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

add a comment |

up vote
3
down vote

favorite

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

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

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

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

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

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

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

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

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

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

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

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

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

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

edited Apr 8 '15 at 13:06

N. F. Taussig

42.8k93254

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

asked Apr 8 '15 at 11:47

Thomas Winkworth

339420

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

add a comment |

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

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

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

add a comment |

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

answered Nov 15 at 19:55

Waffle

457417

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

answered Nov 15 at 19:55

Waffle

457417

add a comment |

up vote
1
down vote

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

answered Nov 15 at 19:55

Waffle

457417

add a comment |

up vote
1
down vote

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

answered Nov 15 at 19:55

Waffle

457417

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

answered Nov 15 at 19:55

Waffle

457417

answered Nov 15 at 19:55

Waffle

457417

answered Nov 15 at 19:55

Waffle

457417

answered Nov 15 at 19:55

Waffle

457417

add a comment |

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