Evaluate - e-xsinxdx-

The correct option is A e^ x2[sinx+cosx]+c Using the identity, #160; ∫e^ax sin(bx + c) = e^axa^2 + b^2× [asin(bx+c) bcos(bx + c)] + c Substituting ...

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Byju's Answer

Standard XII

Mathematics

Geometric Interpretation of Def.Int as Limit of Sum

Evaluate ∫ ...

Question

Evaluate $\int e^{- x} s i n x d x =$

\frac{e^{- x}}{2} [sin x + cos x] + c

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\frac{- e^{- x}}{2} [s i n x + c o s x] + c

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- e^{- x} (s i n x + c o s x) + c

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e^{- x} (s i n x + c o s x) + c

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Solution

The correct option is A $\frac{- e^{- x}}{2} [s i n x + c o s x] + c$
Using the identity,
$\int e^{a x} s i n (b x + c) = \frac{e^{a x}}{a^{2} + b^{2}} \times [a s i n (b x + c) - b c o s (b x + c)] + c$
Substituting the values, with $a = - 1$ and $b = 1$
$\int e^{- x} s i n (x) d x = \frac{e^{- x}}{{(- 1)}^{2} + 1^{2}} \times [- s i n (x) - 1 (c o s x)] + c$
$= - \frac{e^{- x}}{2} [s i n x + c o s x] + c$
Hence, option 'B' is correct.

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Evaluate ∫e−xsinxdx=

The correct option is A −e−x2[sinx+cosx]+cUsing the identity,∫eaxsin(bx+c)=eaxa2+b2×[asin(bx+c)−bcos(bx+c)]+cSubstituting the values, with a=−1 and b=1∫e−xsin(x)dx=e−x(−1)2+12×[−sin(x)−1(cosx)]+c=−e−x2[sinx+cosx]+cHence, option 'B' is correct.

Evaluate $\int e^{- x} s i n x d x =$