We have, I=∫_0^π2e^ xsin xdx Using ∫I.IIdx=I∫IIdx ∫( dIdxIIdx )dx Now, I=∫_0^π2e^ xsin xdx I=sinx∫_0^π2e^ xdx ∫_0^π2[ ( ddxsin x )∫_0^ π2e^ xd ...

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

∫0π /2e-xsin ...

Question

$\int_{0}^{π / 2} e^{- x} s i n x d x$

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Solution

We have,
$I = \int_{0}^{\frac{π}{2}} e^{- x} sin x d x$
Using $\int I . I I d x = I \int I I d x - \int (\frac{d I}{d x} I I d x) d x$
Now,
$I = \int_{0}^{\frac{π}{2}} e^{- x} sin x d x$
$I = sinx \int_{0}^{\frac{π}{2}} e^{- x} d x - \int_{0}^{\frac{π}{2}} ⎡ ⎢ ⎣ (\frac{d}{d x} sin x) \int_{0}^{- \frac{π}{2}} e^{- x} d x ⎤ ⎥ ⎦ d x$
$I = sinx {[- e^{- x}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} (- cos x) (- e^{- x}) d x + C$
$I = sin x ⎡ ⎣ - e^{- \frac{π}{2}} + e^{0} ⎤ ⎦ - \int_{0}^{\frac{π}{2}} e^{- x} cos x d x + C$
$I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ sin x - ⎡ ⎢ ⎣ cos x \int_{0}^{\frac{π}{2}} e^{- x} d x - \int_{0}^{\frac{π}{2}} ⎡ ⎢ ⎣ (\frac{d}{d x} cos x) \int_{0}^{\frac{π}{2}} e^{- x} d x ⎤ ⎥ ⎦ d x ⎤ ⎥ ⎦ + C$
$I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ sin x - ⎡ ⎢ ⎣ cos x {(- e^{- x})}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} sin x (- e^{- x}) d x ⎤ ⎥ ⎦ + C$
$I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ sin x - ⎡ ⎣ cos x {(- e^{- x})}_{0}^{\frac{π}{2}} + I ⎤ ⎦ + C$
$I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ sin x - cos x {(- e^{- x})}_{0}^{\frac{π}{2}} - I + C$
$2 I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ sin x - cos x ⎡ ⎣ - e^{- \frac{π}{2}} + e^{0} ⎤ ⎦ + C$
$2 I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ sin x - ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ cos x + C$
$2 I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ (sin x - cos x) + C$
$I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ \frac{(sin x - cos x)}{2} + \frac{C}{2}$
$I = ⎛ ⎝ 1 - e^{- \frac{π}{2}} ⎞ ⎠ \frac{(sin x - cos x)}{2} + C^{'}$
Hence, this is the answer.

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