MHT CET · Maths · Indefinite Integration
\(\int e^{\tan x}\left(\sec ^{2} x+\sec ^{3} x \sin x\right) d x\) is equal to
- A \(\sec x e^{\tan x}+c\)
- B \(\tan x e^{\tan x}+c\)
- C \(e^{\tan x}+\tan x+c\)
- D \((1+\tan x) e^{\tan x}+c\)
Answer & Solution
Correct Answer
(B) \(\tan x e^{\tan x}+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int e^{\tan x}\left(\sec ^{2} x+\sec ^{3} x \sin x\right) d x\)
\(=\int e^{\tan x}(1+\tan x) \sec ^{2} x d x\)
Put \(\tan x=t\)
\(\Rightarrow \sec ^{2} x d x=d t\)
\(I=\int e^{t}(1+t) d t\)
\(=e^{t}+\int t e^{t} d t\)
\(=e^{t}+t e^{t}-e^{t}+c\)
\(=t e^{t}+c\)
\(=\tan x e^{\tan x}+c\)
\(=\int e^{\tan x}(1+\tan x) \sec ^{2} x d x\)
Put \(\tan x=t\)
\(\Rightarrow \sec ^{2} x d x=d t\)
\(I=\int e^{t}(1+t) d t\)
\(=e^{t}+\int t e^{t} d t\)
\(=e^{t}+t e^{t}-e^{t}+c\)
\(=t e^{t}+c\)
\(=\tan x e^{\tan x}+c\)
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