trig calculus qs (1 Viewer)

[xXnukerrrXx]

Qs:


Change tan x = sin x/cos x


show that if you integrate tan x from 0 to pi/4 the answer would be 1/2 ln(2)

 

[Nooblet94]

<a href="http://www.codecogs.com/eqnedit.php?latex=\int^{\frac{\pi}{4}}_0 \tan xdx=\int^{\frac{\pi}{4}}_0 \frac{\sin x}{\cos x}dx=-\int^{\frac{\pi}{4}}_0 \frac{-\sin x}{\cos x}dx \\ \textrm{Let}~u=\cos x\\ \textrm{When}~x=\frac{\pi}{4}~u=\frac{1}{\sqrt{2}}\\ \textrm{When}~x=0~u=1\\ ~\\ \textrm{Now, we know that}~\frac{d}{dx} \cos x=-\sin x\\ \therefore du=-\sin xdx\\~\\ \textrm{Upon substituting in for u we get}\\ \int^{\frac{\pi}{4}}_0 \tan xdx\\ =-\int^{\frac{1}{\sqrt{2}}}_{1}\frac{1}{u}du\\ =-\left[ \ln u\right]^{\frac{1}{\sqrt{2}}}_{1}\\ =-\ln(\frac{1}{\sqrt{2}}) @plus; \ln(1)\\ =\ln(\sqrt{2}) =\ln(2^{\frac{1}{2}}) =\frac{1}{2}\ln2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int^{\frac{\pi}{4}}_0 \tan xdx=\int^{\frac{\pi}{4}}_0 \frac{\sin x}{\cos x}dx=-\int^{\frac{\pi}{4}}_0 \frac{-\sin x}{\cos x}dx \\ \textrm{Let}~u=\cos x\\ \textrm{When}~x=\frac{\pi}{4}~u=\frac{1}{\sqrt{2}}\\ \textrm{When}~x=0~u=1\\ ~\\ \textrm{Now, we know that}~\frac{d}{dx} \cos x=-\sin x\\ \therefore du=-\sin xdx\\~\\ \textrm{Upon substituting in for u we get}\\ \int^{\frac{\pi}{4}}_0 \tan xdx\\ =-\int^{\frac{1}{\sqrt{2}}}_{1}\frac{1}{u}du\\ =-\left[ \ln u\right]^{\frac{1}{\sqrt{2}}}_{1}\\ =-\ln(\frac{1}{\sqrt{2}}) + \ln(1)\\ =\ln(\sqrt{2}) =\ln(2^{\frac{1}{2}}) =\frac{1}{2}\ln2" title="\int^{\frac{\pi}{4}}_0 \tan xdx=\int^{\frac{\pi}{4}}_0 \frac{\sin x}{\cos x}dx=-\int^{\frac{\pi}{4}}_0 \frac{-\sin x}{\cos x}dx \\ \textrm{Let}~u=\cos x\\ \textrm{When}~x=\frac{\pi}{4}~u=\frac{1}{\sqrt{2}}\\ \textrm{When}~x=0~u=1\\ ~\\ \textrm{Now, we know that}~\frac{d}{dx} \cos x=-\sin x\\ \therefore du=-\sin xdx\\~\\ \textrm{Upon substituting in for u we get}\\ \int^{\frac{\pi}{4}}_0 \tan xdx\\ =-\int^{\frac{1}{\sqrt{2}}}_{1}\frac{1}{u}du\\ =-\left[ \ln u\right]^{\frac{1}{\sqrt{2}}}_{1}\\ =-\ln(\frac{1}{\sqrt{2}}) + \ln(1)\\ =\ln(\sqrt{2}) =\ln(2^{\frac{1}{2}}) =\frac{1}{2}\ln2" /></a>

 

[_deloso]

Qs:


Change tan x = sin x/cos x


show that if you integrate tan x from 0 to pi/4 the answer would be 1/2 ln(2)

integrate sin x/cos x: it becomes - ln cos x (differentiate a ln function: f'(x)/f(x). If you differntiate -ln cos x, it becomes - f'(x)/f(x) = - -sin x/cos x = sin x/ cos x, thus integral of sin x/ cos x is - ln cos x
now you sub in the bounds (0 and pi/4 for x)
= -(ln cos pi/4 -ln cos 0)
using exact values
= -ln 1/sqr2 + ln 1
ln1 is 0
thus
= -ln1/sqrt2
=-ln 2 to the power of -(1/2) --> 1/sqrt2 is the same as 2 to the power of -(1/2) using indices rule
use the power rule of log where you can bring down the power in a ln function
= -(1/2) times -ln2
=1/2 ln 2
as req.

probably a little bit messy if you dont know your log rules and indices rules