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Question: \(\int_{}^{}e^{ax}\).sin bx dx is \(\frac{e^{ax}}{l}\) (a sin bx – b cos bx) + c. Then l is –...

∫eax\int_{}^{}e^{ax}.sin bx dx is eaxl\frac{e^{ax}}{l} (a sin bx – b cos bx) + c. Then l is –

A

a2 + b2

B

(a2+b2)\sqrt{(a^{2} + b^{2})}

C

(a2 + b2)2

D

None of these

Answer

a2 + b2

Explanation

Solution

I = . dx = sin bx . – cos bx . dx

= – . cos bx dx

= –

= – . cos bx –

Ž I = (a sin bx – b cos bx)

Ž I = (a sin bx – b cos bx) + c

Hence l = (a2 + b2).