- e x x-log x-tan x d x

Let #160;I= #8747;exsecx+logsecx+tanxdxHere, #160;f(x)=logsecx+tanx #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160 ...

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Byju's Answer

Standard XII

Mathematics

Arithmetic Mean

∫ e x x+log x...

Question

$\int e^{x} [\sec x + \log (\sec x + \tan x)] d x$

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Solution

$Let I = \int e^{x} [\sec x + \log (\sec x + \tan x)] d x Here, f (x) = \log (\sec x + \tan x) Put e^{x} f (x) = t \Rightarrow f' (x) = \sec x let e^{x} \log (\sec x + \tan x) = t Diff both sides w . r . t x e^{x} \log (\sec x + \tan x) + e^{x} \frac{1}{\sec x + \tan x} (\sec x + \tan x + \sec^{2} x) = \frac{d t}{d x} \Rightarrow [e^{x} \log (\sec x + \tan x) + e^{x} (\sec x)] d x = d t \Rightarrow e^{x} [\sec x + \log (\sec x + \tan x)] d x = d t ∴ \int e^{x} [\sec x + \log (\sec x + \tan x)] d x = \int d t = t + C = e^{x} \log (\sec x + \tan x) + C$

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∫ex sec x+ log sec x+tan x dx

Let I=∫exsecx+logsecx+tanxdxHere, f(x)=logsecx+tanx Put exf(x)=t⇒f'(x)=secx let exlogsecx+tanx=tDiff both sides w.r.t xexlogsecx+tanx+ex1secx+tanxsecx+tanx+sec2x=dtdx⇒exlogsecx+tanx+exsecxdx=dt⇒exsecx+logsecx+tanxdx=dt∴∫exsecx+logsecx+tanxdx=∫dt=t+C=exlogsecx+tanx+C

$\int e^{x} [\sec x + \log (\sec x + \tan x)] d x$