dstefan
03-22-2008, 01:04 PM
Chapter 1, Problem 3:
Show that (\tan x)' = \frac{1}{(\cos x)^2}
and
\int \frac{1}{1+x^2} ~dx ~=~ \arctan(x) ~+~ C
Solution:
Quotient Rule:
(\tan x)' = \left(\frac{\sin x}{\cos x}\right)' = \frac{(\sin x)' \cos x - \sin x (\cos x)'}{(\cos x)^2} = \frac{(\cos x)^2 + (\sin x)^2}{(\cos x)^2} = \frac{1}{(\cos x)^2}
If we know that \int \frac{1}{1+x^2} ~dx is \arctan(x), all we need to show is that (\arctan(x))' = \frac{1}{1+x^2}.
Let f(x) = \tan x. Then \arctan(x) = f^{-1}(x). We know that \left( f^{-1}(x) \right)' ~=~ \frac{1}{f'(f^{-1}(x))}.
Recall that f'(x) = (\tan x)' = \frac{1}{(\cos x)^2}. Then, (\arctan(x))' = (\cos(f^{-1}(x)))^2 = (\cos(\arctan(x)))^2.
Let \alpha = \arctan(x). Then \tan(\alpha) = x. It is easy to see that x^2 +1 = \frac{1}{(\cos(\alpha))^2} and thus (\cos(\arctan(x)))^2 = (\cos(\alpha))^2 = \frac{1}{x^2+1}.
We conclude that (\arctan(x))' = \frac{1}{x^2+1}, and therefore that \int \frac{1}{1+x^2} ~dx ~=~ \arctan(x) ~+~ C.
If we do not know what the antiderivative of \frac{1}{1+x^2} is \arctan(x), the substitution x = \tan \left(\frac{z}{2}\right) can be used.
Note that dx = \frac{d}{dz} \tan\left(\frac{z}{2}\right) dz = \frac{1}{2(\cos \left(\frac{z}{2}\right))^2} dz. Then
\int \frac{1}{1+x^2} ~dx = \int \frac{1}{1+(\tan \left(\frac{z}{2}\right))^2} \cdot \frac{1}{2(\cos \left(\frac{z}{2}\right))^2} dz = \int (\cos \left(\frac{z}{2}\right))^2 \frac{1}{2(\cos \left(\frac{z}{2}\right))^2} dz = \frac{z}{2} = \arctan(x) ~+~ C.
Show that (\tan x)' = \frac{1}{(\cos x)^2}
and
\int \frac{1}{1+x^2} ~dx ~=~ \arctan(x) ~+~ C
Solution:
Quotient Rule:
(\tan x)' = \left(\frac{\sin x}{\cos x}\right)' = \frac{(\sin x)' \cos x - \sin x (\cos x)'}{(\cos x)^2} = \frac{(\cos x)^2 + (\sin x)^2}{(\cos x)^2} = \frac{1}{(\cos x)^2}
If we know that \int \frac{1}{1+x^2} ~dx is \arctan(x), all we need to show is that (\arctan(x))' = \frac{1}{1+x^2}.
Let f(x) = \tan x. Then \arctan(x) = f^{-1}(x). We know that \left( f^{-1}(x) \right)' ~=~ \frac{1}{f'(f^{-1}(x))}.
Recall that f'(x) = (\tan x)' = \frac{1}{(\cos x)^2}. Then, (\arctan(x))' = (\cos(f^{-1}(x)))^2 = (\cos(\arctan(x)))^2.
Let \alpha = \arctan(x). Then \tan(\alpha) = x. It is easy to see that x^2 +1 = \frac{1}{(\cos(\alpha))^2} and thus (\cos(\arctan(x)))^2 = (\cos(\alpha))^2 = \frac{1}{x^2+1}.
We conclude that (\arctan(x))' = \frac{1}{x^2+1}, and therefore that \int \frac{1}{1+x^2} ~dx ~=~ \arctan(x) ~+~ C.
If we do not know what the antiderivative of \frac{1}{1+x^2} is \arctan(x), the substitution x = \tan \left(\frac{z}{2}\right) can be used.
Note that dx = \frac{d}{dz} \tan\left(\frac{z}{2}\right) dz = \frac{1}{2(\cos \left(\frac{z}{2}\right))^2} dz. Then
\int \frac{1}{1+x^2} ~dx = \int \frac{1}{1+(\tan \left(\frac{z}{2}\right))^2} \cdot \frac{1}{2(\cos \left(\frac{z}{2}\right))^2} dz = \int (\cos \left(\frac{z}{2}\right))^2 \frac{1}{2(\cos \left(\frac{z}{2}\right))^2} dz = \frac{z}{2} = \arctan(x) ~+~ C.