Evaluate the integral -0-e-xsin bxxdx

Let nbsp; nbsp; I=∫_0^∞e^ xsin bxxdx dIdb=∫_0^∞∂∂ b ( e^ xsin bxx )dx=∫_0^∞e^ xxcos bxxdx =∫_0^∞e^ xcos bx dx We know than ∫ e^axcos bx =e^axa^2+b^2(a cos b ...

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Engineering Mathematics

Reduction Formulae and Trigonometric System

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Question

Evaluate the integral $\int_{0}^{\infty} \frac{e^{- x} s i n b x}{x} d x$

c o t^{- 1} b

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t a n^{- 1} b

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s e c^{- 1} b

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c o s e c^{- 1} b

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Solution

The correct option is B $t a n^{- 1} b$
Let $I = \int_{0}^{\infty} \frac{e^{- x} s i n b x}{x} d x$

$\frac{d I}{d b} = \int_{0}^{\infty} \frac{\partial}{\partial b} (\frac{e^{- x} s i n b x}{x}) d x = \int_{0}^{\infty} \frac{e^{- x} x c o s b x}{x} d x$

$= \int_{0}^{\infty} e^{- x} c o s b x d x$
We know than
$\int e^{a x} c o s b x = \frac{e^{a x}}{a^{2} + b^{2}} (a c o s b x + b s i n b x)$

$\int_{0}^{\infty} e^{- x} c o s b x d x = {[\frac{e^{- x}}{1 + b^{2}} [- c o s b x + b s i n b x]]}_{0}^{\infty}$

$\frac{d I}{d b} = \frac{1}{1 + b^{2}}$
Integrating both sides, $I = t a n^{- 1} b$

Suggest Corrections

Evaluate the integral ∫∞0e−xsin bxxdx

The correct option is B tan−1bLet I=∫∞0e−xsin bxxdx dIdb=∫∞0∂∂b(e−xsin bxx)dx=∫∞0e−xxcos bxxdx =∫∞0e−xcos bx dx We know than ∫eaxcos bx=eaxa2+b2(a cos bx+b sin bx) ∫∞0e−xcosbxdx=[e−x1+b2[−cosbx+b sin bx]]∞0 dIdb=11+b2 Integrating both sides, I=tan−1b

Evaluate the integral $\int_{0}^{\infty} \frac{e^{- x} s i n b x}{x} d x$