Question

∫ eˣ sin x cos x dx=.........+c ,Select correct option from the given options.
(a) eˣ/2√5 cos(2x-tan⁻¹2)
(b) eˣ/2√5 sin(2x-tan⁻¹2)
(c) e²/2√5 sin(2x+tan⁻¹2)
(d) e²ˣ/2√5 sin(2x+π-tan⁻¹2)

Answer 1

answer : option (b)

explanation : it is based on concept,

$\int{e^{ax}sinbx}\,dx=\frac{e^{ax}}{a^2+b^2}(asinb-bcosbx)+C$

let's arrange given integrant,

∫e^x sinx cosx dx

= ∫e^x {1/2(2sinx cosx)} dx

we know, 2sinθcosθ = sin2θ

so, 2sinx cosx = sin2x

now, ∫e^x {1/2(2sinx cosx)} dx =1/2 ∫e^x sin2x

here, a = 1 and b = 2

from above concept,

1/2 ∫e^x sin2x = 1/2 [ e^x/(1² + 2²) (1 × sin2x - 2cos2x )]

= 1/2 [1/5 e^x(sin2x - 2cos2x )] + C

= 1/10 [e^x (sin2x - 2cos2x )] + C

we know, asinθ ± bcosθ = √(a² + b²)sin(θ±tan^-1(b/a))

so, sin2x - 2cos2x = √(1² + 2²)sin{2x - tan^-1(2/1)} = √5sin(2x - tan-¹2)

then, 1/10 [e^x √5 sin(2x - tan-¹2)] +C

= 1/2√5 [e^x sin(2x - tan-¹2 )] + C

hence, option (b) is correct