-0- e - ax cos bxdx -A- None of theseB- a-a2-b2C- 10-a2-b2D- b-a2-b2

The correct option is B a/a^2+b^2Let I=∫_0^∞e^ ax cos bx dx = [ ( 1/acosbx.e^ ax ) |^∞_0 .+∫_0^∞b/asinbxe^ axdx] =1/a b^2/a^2I ⇒ ( 1+b^2/a^2 )I=1/a ⇒ I=a/a^ ...

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Byju's Answer

Standard XII

Mathematics

Property 3

∫0∞ e - ax co...

Question

$\int_{0}^{\infty} e^{- a x} c o s b x d x =$

\frac{b}{a^{2} + b^{2}}

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\frac{b}{a^{2} - b^{2}}

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\frac{a}{a^{2} + b^{2}}

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Solution

The correct option is C $\frac{a}{a^{2} + b^{2}}$
$L e t I = \int_{0}^{\infty} e^{- a x} c o s b x d x$
$= {[(\frac{- 1}{a} c o s b x . e^{- a x}) ∣ ∣}_{0}^{\infty} + \int_{0}^{\infty} \frac{b}{a} s i n b x e^{- a x} d x]$
$= \frac{1}{a} - \frac{b^{2}}{a^{2}} I \Rightarrow (1 + \frac{b^{2}}{a^{2}}) I = \frac{1}{a}$
$\Rightarrow I = \frac{a}{a^{2} + b^{2}}$

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∫∞0e−axcos bxdx=

The correct option is C aa2+b2Let I=∫∞0e−ax cos bx dx =[(−1acosbx.e−ax)∣∣∞0+∫∞0basinbxe−axdx] =1a−b2a2I⇒(1+b2a2)I=1a ⇒I=aa2+b2

$\int_{0}^{\infty} e^{- a x} c o s b x d x =$