Question

e^sinx sin2xdx integral

Answer 1

i hope this will help you

Answer 2

∫ eˢⁱⁿˣ sin2x dx = 2 sinx (eˢⁱⁿˣ - 1) + C

Step-by-step explanation:

We have to find the integral of ∫ eˢⁱⁿˣ sin2x dx

We use the transformation, sinx = z to reduce the integrand to a simple form.

Here, sinx = z

Taking differentials in both sides, we get

d (sinx) = d (z)

or, cosx dx = dz ..... (1)

Now, ∫ eˢⁱⁿˣ sin2x dx

= ∫ eˢⁱⁿˣ . 2 sinx cosx dx [ ∵ sin2x = 2 sinx cosx ]

= 2 ∫ eᶻ . z dz [ by (1) ]

= 2 [ z ∫ eᶻ dz - ∫ {d/dz (z) . ∫ eᶻ dz} dz + C

where C is the constant of integration

= 2 (z eᶻ - ∫ eᶻ dz) + C

= 2 (z eᶻ - eᶻ) + C

= 2 (z - 1) eᶻ + C

= 2 (sinx - 1) eˢⁱⁿˣ + C [ ∵ z = sinx ]

∴ ∫ eˢⁱⁿˣ sin2x dx = 2 eˢⁱⁿˣ (sinx - 1) + C,

where C is constant of integration

Rule:

∫ uv dx = u ∫ v dx - ∫ [du/dx . ∫ v dx] dx

where u, v are functions of x