Q111 / 180 · KCET 2011 · Mathematics

$\int e^{x}\left[\frac{\sin x+\cos x}{1-\sin ^{2} x}\right] d x$ is

  • A
    $\left(e^{x} \cdot \operatorname{cosec} x\right)+C$
  • B
    $e^{x} \cot x+C$
  • Correct answer
    C
    $\left(e^{x} \cdot \sec x\right)+C$
  • D
    $e^{x} \tan x+C$

Solution

$I=\int e^{x}\left(\frac{\sin x+\cos x}{1-\sin ^{2} x}\right) d x$ $I=\int e^{x}\left(\frac{\sin x+\cos x}{\cos ^{2} x}\right) d x$ $I=\int e^{x} \tan x \cdot \sec x d x+\int e^{x} \cdot \sec x d x$ $I=\int e^{x} \tan x \cdot \sec x d x+\left\{\sec x \cdot e^{x}\right.$ $\left.I=\int \sec x \cdot \tan x e^{x} d x\right\}$ $I=e^{x} \tan x \cdot \sec x d x+e^{x} \sec x$ $\quad-\int e^{x} \tan x \cdot \sec x d x$

Hence, the correct answer is option: 'C'.

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