definite integral (1 Viewer)

cheesynooby · Sunday at 5:01 PM

there's this integral where the answers do t-subs but I'm wondering whether multiplying both the numerator and denominator by sin(theta) - 1 as shown below (with many lines of working skipped) would work.
$\int_{0}^{pi/2}{\frac{1}{sin\theta+1}}^{}{d\theta} = \int_{0}^{pi/2}{tan{\theta}sec\theta-sec^2{\theta}}^{}{d\theta}$
which gives the primitive secθ - tanθ, but it is undefined at pi/2.
so my question is if you can do anything with my method and how you could tell when you need to use a t-sub?

this is probably extension 2 content but the ideas are probably somewhat relevant to extension 1 so im posting it here
(sorry idk how to make it show π instead of pi, also i believe this is a question from the 2024 x2 HSC and the answer is 1)

tywebb · Sunday at 5:17 PM

cheesynooby said:
there's this integral where the answers do t-subs but I'm wondering whether multiplying both the numerator and denominator by sin(theta) - 1 as shown below (with many lines of working skipped) would work.
$\int_{0}^{pi/2}{\frac{1}{sin\theta+1}}^{}{d\theta} = \int_{0}^{pi/2}{tan{\theta}sec\theta-sec^2{\theta}}^{}{d\theta}$
which gives the primitive secθ - tanθ, but it is undefined at pi/2.
so my question is if you can do anything with my method and how you could tell when you need to use a t-sub?

this is probably extension 2 content but the ideas are probably somewhat relevant to extension 1 so im posting it here
(sorry idk how to make it show π instead of pi, also i believe this is a question from the 2024 x2 HSC and the answer is 1)

It's the other way around

$\frac{1}{\sin\theta+1}=\sec^2\theta-\tan\theta\sec\theta$

Answer is 1 and

$\int_0^\frac{\pi}{2}(\sec^2\theta-\tan\theta\sec\theta)\ \!d\theta=1$

so method is ok, but just do it the other way around

cheesynooby · Sunday at 5:30 PM

tywebb said:
It's the other way around

$\frac{1}{\sin\theta+1}=\sec^2\theta-\tan\theta\sec\theta$

Answer is 1 and

$\int_0^\frac{\pi}{2}(\sec^2\theta-\tan\theta\sec\theta)\ \!d\theta=1$

so method is ok, but just do it the other way around

ok but how do you evaluate the integral since pi/2 is undefined for secx and tanx

tywebb · Sunday at 5:33 PM

here another way

$\int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}\frac{1}{2}\csc^2\left(\frac{\theta}{2}+\frac{\pi}{4}\right)\ \!d\theta=1$

tywebb · Sunday at 5:38 PM

cheesynooby said:
ok but how do you evaluate the integral since pi/2 is undefined for secx and tanx

u can do limit

$\lim_{\theta\rightarrow\frac{\pi}{2}}(\tan\theta-\sec\theta)=0$

tywebb · Sunday at 5:45 PM

technically in the context of this integral one only needs to do the left-sided limit

$\lim_{\theta\rightarrow\(\frac{\pi}{2})^-}(\tan\theta-\sec\theta)=0$

but it works from both sides anyway

look at graph of y=tan x-sec x

u can see it approaches 0 as x->pi/2

cheesynooby · Sunday at 5:57 PM

tywebb said:
technically in the context of this integral one only needs to do the left-sided limit

$\lim_{\theta\rightarrow\(\frac{\pi}{2})^-}(\tan\theta-\sec\theta)=0$

but it works from both sides anyway

look at graph of y=tan x-sec x

u can see it approaches 0 as x->pi/2
View attachment 46911

ok i get that the limit is 0 but how am i supposed to work that out without a graphing calculator/is there a way to remove the need for limits

tywebb · Sunday at 5:58 PM

u know why this?

i show u now

$\begin{aligned}\lim_{\theta\rightarrow\frac{\pi}{2}}(\tan\theta-\sec\theta) &=\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\sin\theta-1}{\cos\theta}\\ &=\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\frac{d}{d\theta}(\sin\theta-1)}{\frac{d}{d\theta}\cos\theta}\\ &=\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\cos\theta}{-\sin\theta}\\ &=\frac{0}{-1}\\ &=0 \end{aligned}$

good?

tywebb · Sunday at 6:06 PM

now we redo the original question.

i don't think at this stage we need to give up on your method. it works! but just other way around.

$\int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}(\sec^2\theta-\sec\theta\tan\theta)\ \!d\theta=\lim_{\alpha\rightarrow(\frac{\pi}{2})^-}[\tan\theta-\sec\theta]_0^\alpha=0-(0-1)=1$

good?

tywebb · Sunday at 6:51 PM

cheesynooby said:
is there a way to remove the need for limits

yeah u can actually

did u know that

$\tan\theta-\sec\theta=\sin\left(\frac{\theta}{2}-\frac{\pi}{4}\right)\csc\left(\frac{\theta}{2}+\frac{\pi}{4}\right)$

whereupon we may do this one without the limits

$\int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}(\sec^2\theta-\sec\theta\tan\theta)\ \!d\theta=[\tan\theta-\sec\theta]_0^\frac{\pi}{2}=\left[\sin\left(\frac{\theta}{2}-\frac{\pi}{4}\right)\csc\left(\frac{\theta}{2}+\frac{\pi}{4}\right)\right]_0^\frac{\pi}{2}=0\times1-\frac{-1}{\sqrt2}\times\sqrt2=1$

tywebb · Sunday at 7:18 PM

alternatively, u know

$\tan\theta-\sec\theta=1-\frac{2}{\tan\frac{\theta}{2}+1}$

and so

$\int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}(\sec^2\theta-\sec\theta\tan\theta)\ \!d\theta=[\tan\theta-\sec\theta]_0^\frac{\pi}{2}=\left[1-\frac{2}{\tan\frac{\theta}{2}+1}\right]_0^\frac{\pi}{2}=1-1-(1-2)=1$

i think limit method is a bit neater than these 2 methods though

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definite integral (1 Viewer)

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