13. $\\frac{du}{dx}=x\\sec^{2}x$ and $u = 1$ when $x = 0$
13. $\\frac{du}{dx}=x\\sec^{2}x$ and $u = 1$ when $x = 0$
Answer
Explanation:
Step1: Integrate both sides
$$u=\int x\sec^{2}x dx$$ Use integration by parts. Let (v = x), (dw=\sec^{2}x dx). Then (dv = dx), (w=\tan x). By the integration - by - parts formula (\int vdw=vw-\int wdv), we have: $$u=x\tan x-\int\tan xdx$$
Step2: Integrate (\int\tan xdx)
Since (\int\tan xdx=\int\frac{\sin x}{\cos x}dx). Let (t = \cos x), (dt=-\sin xdx). Then (\int\frac{\sin x}{\cos x}dx=-\int\frac{dt}{t}=-\ln|\cos x| + C) So (u=x\tan x+\ln|\cos x|+C)
Step3: Use the initial condition
When (x = 0), (u = 1). Substitute (x = 0) and (u = 1) into (u=x\tan x+\ln|\cos x|+C) (1=0\times\tan0+\ln|\cos0|+C). Since (\cos0 = 1) and (\ln1=0), then (C = 1)
Answer:
(u=x\tan x+\ln|\cos x| + 1)