Posted by **kchik** on Tuesday, December 4, 2012 at 12:09am.

INTEGRATION BY PARTS...

1) e^5t sin 3t dt

2) xe^3x dx

3) x sec^2 x dx

**anybody can help me for this question?

- calculus -
**Steve**, Tuesday, December 4, 2012 at 4:40am
if we let A be our integral, we have

A = ∫e^5t sin3t dt

u=e^5t

du = 5e^5t dt

dv = sin3t dt

v = -1/3 cos3t

A = uv - ∫v du

= -1/3 e^5t cos3t + 5/3 ∫e^5t cos3t dt

u=e^5t

du = 5e^5t dt

dv = cos3t dt

v = 1/3 sin3t

A = -1/3 e^5t cos3t + 5/3(1/3 e^5t sin3t - 5/3 ∫e^5t sin3t dt)

= -1/3 e^5t cos3t + 5/9 e^5t sin3t - 25/9 A

34/9 A = 1/9 e^5t (-3cos3t + 5sin3t)

A = 1/34 e^5t (-3cos3t + 5sin3t)

-----------------------------------

A = ∫xe^3x dx

u = x

du = dx

dv = e^3x dx

v = 1/3 e^3x

A = uv - ∫v du

= 1/3 xe^3x - 1/3 ∫e^3x dx

= 1/3 xe^3x - 1/9 e^3x

= 1/9 e^3x (3x-1)

------------------------

A = ∫x sec^2 x dx

u = x

du = dx

dv = sec^2 x dx

v = tan x

A = uv - ∫v du

= x tanx - ∫tanx dx

= x tanx - ∫sinx/cosx dx

= x tanx + log cosx

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