KCET · Maths · Indefinite Integration
\(\int e^{x}\left[\frac{\sin x+\cos x}{1-\sin ^{2} x}\right] d x\) is
- A \(\left(e^{x} \cdot \operatorname{cosec} x\right)+C\)
- B \(e^{x} \cot x+C\)
- C \(\left(e^{x} \cdot \sec x\right)+C\)
- D \(e^{x} \tan x+C\)
Answer & Solution
Correct Answer
(C) \(\left(e^{x} \cdot \sec x\right)+C\)
Step-by-step Solution
Detailed explanation
\(I=\int e^{x}\left(\frac{\sin x+\cos x}{1-\sin ^{2} x}\right) d x\)
\(I=\int e^{x}\left(\frac{\sin x+\cos x}{\cos ^{2} x}\right) d x\)
\(I=\int e^{x} \tan x \cdot \sec x d x+\int e^{x} \cdot \sec x d x\)
\(I=\int e^{x} \tan x \cdot \sec x d x+\left\{\sec x \cdot e^{x}\right.\)
\(\left.I=\int \sec x \cdot \tan x e^{x} d x\right\}\)
\(I=e^{x} \tan x \cdot \sec x d x+e^{x} \sec x\)
\(\quad-\int e^{x} \tan x \cdot \sec x d x\)
\(I=\int e^{x}\left(\frac{\sin x+\cos x}{\cos ^{2} x}\right) d x\)
\(I=\int e^{x} \tan x \cdot \sec x d x+\int e^{x} \cdot \sec x d x\)
\(I=\int e^{x} \tan x \cdot \sec x d x+\left\{\sec x \cdot e^{x}\right.\)
\(\left.I=\int \sec x \cdot \tan x e^{x} d x\right\}\)
\(I=e^{x} \tan x \cdot \sec x d x+e^{x} \sec x\)
\(\quad-\int e^{x} \tan x \cdot \sec x d x\)
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