How can we integrate sec x? Please give me only hint.
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Originally Posted by roshanhero How can we integrate sec x? Please give me only hint. Let $\displaystyle x=2\arctan(z)$. That is the most fundamental and fool-proof way. There are cleverer ones.
another way is to multiply and divide sec(x) by (secx +tan(x) to obtain [sec^2(x) +sec(x)tan(x)]/[sec(x)+tan(x)] Then let u = sec(x) +tan(x) du = [ sec^2(x) +sec(x)tan(x)]dx you end up with integral[du/u] = ln(u)
Multiply top and bottom by $\displaystyle \sec x+\tan x.$ (Got beaten.)
multiply top and bottom by $\displaystyle \cos x$.
Don't really see what multiplying top and bottom by cos(x) does for you??
Originally Posted by Calculus26 Don't really see what multiplying top and bottom by cos(x) does for you?? After that, substitute $\displaystyle \cos ^2x$ with $\displaystyle 1-\sin ^2x$, then seperate the fraction into two parts, I think you can take it from here now.
Originally Posted by Calculus26 Don't really see what multiplying top and bottom by cos(x) does for you?? $\displaystyle \int\frac{dx}{\cos(x)}=\int\frac{\cos(x)}{\cos^2(x )}dx=\int\frac{\cos(x)}{1-\sin^2(x)}dx$ Let $\displaystyle \sin(x)=z\implies dz=\cos(x)\text{ }dx$ So our integral becomes $\displaystyle \int\frac{dz}{1-z^2}$
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