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Differentiating Trig

Part one : show that 1/sinxcosx=sec^2x/tanx

Did part one fine but part 2: Hence find the exact value of S(integralsign) 1/sinxcosx beteen 1/3pi and1/4pi

Help im totally lost
(edited 10 years ago)
Does the RHS look like kf'(x)f(x)?

Try differentiating (1/2)tan^2(x)

Or maybe you should try integrating by parts. There's more than one way around this.
Original post by Angryification
Does the RHS look like kf'(x)f(x)?

Try differentiating (1/2)tan^2(x)

Or maybe you should try integrating by parts. There's more than one way around this.


I think it would be easiest to note that 12sin⁡2x=sin⁡xcos⁡x\frac{1}{2} \sin 2x = \sin x \cos x and reach for the formula book.
Original post by Mr M
I think it would be easiest to note that 12sin⁡2x=sin⁡xcos⁡x\frac{1}{2} \sin 2x = \sin x \cos x and reach for the formula book.

That's what I'd do unless it said Hence.
Original post by emma201295
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Original post by Mr M
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Original post by keromedic
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Unparseable latex formula:

\displaystyle \begin{aligned} I & = \int \dfrac{dx}{\sin x \cos x} \\ & = \int \dfrac{\sec^2 x}{\tan x}\ dx


It's worth noticing that the integral is of the form ∫f′(x)f(x) dx\displaystyle \int \dfrac{f'(x)}{f(x)}\ dx
Original post by Khallil
Unparseable latex formula:

\displaystyle \begin{aligned} I & = \int \dfrac{dx}{\sin x \cos x} \\ & = \int \dfrac{\sec^2 x}{\tan x}\ dx


It's worth noticing that the integral is of the form ∫f′(x)f(x) dx\displaystyle \int \dfrac{f'(x)}{f(x)}\ dx


Yes we know.

Edit: I misread post 2 though - I thought this had already been pointed out.
(edited 10 years ago)
Original post by Mr M
Yes we know. See post 2.


Original post by Angryification

#2



I've seen post 2. The last part is incorrect.

Angryification is referring to the integral of the product of tan⁡x\tan x and sec⁡2x\sec^2 x, whereas I'm referring to the quotient in question.
Original post by Khallil
I've seen post 2. The last part is incorrect.

Angryification is referring to the integral of the product of tan⁡x\tan x and sec⁡x\sec x whereas I'm referring to the quotient in question.


Reread my last post.
Original post by Mr M
Reread my last post.


No worries, we all make mistakes. :smile:
Original post by Khallil
No worries, we all make mistakes. :smile:


My contribution was supposed to be an alternative to the *obvious* substitution or recognition of a logarithmic integral. I like students to instantly recognise the product sin x cos x.
Reply 10
Avatar for em.d_4
em.d_4
OP
17
Original post by Mr M
I think it would be easiest to note that 12sin⁡2x=sin⁡xcos⁡x\frac{1}{2} \sin 2x = \sin x \cos x and reach for the formula book.

Thanks but how would you use this t then integrate?
Original post by emma201295
Thanks but how would you use this t then integrate?


So ∫1sin⁡xcos⁡x dx=2∫csc⁡2x dx\displaystyle \int \frac{1}{\sin x \cos x} \, dx = 2 \int \csc 2x \, dx

You can look this up in your formula book.
Reply 12
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em.d_4
OP
17
Thanks for your help! Got there in the end :smile:

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