Integration By Parts

Answer 1

Let u = e^ax and dv = cos(bx)dx. Then du = ae^ax and v = sin(bx)/b. Then, we have:

∫ e^axcos(bx)dx = e^axsin(bx)/b - (a/b) ∫e^axsin(bx)dx

Now, let u = e^ax and dv = sin(bx)dx. Then du = ae^ax and v = -cos(bx)/b. Then, we have:

∫ e^axcos(bx)dx = e^axsin(bx)/b +ae^axcos(bx)/b² - (a²/b²) ∫ e^axcos(bx)dx

Now, we need to solve the last equation for ∫ e^axcos(bx)dx. We have:

[1+(a²/b²)] ∫ e^axcos(bx)dx = e^ax(sin(bx)/b + acos(bx)/b²) = e^ax(acos(bx) + bsin(bx))/b²

∫ e^axcos(bx)dx = e^ax(acos(bx) + bsin(bx))/[b²[1+(a²/b²)]] = e^ax(acos(bx) + bsin(bx))/(a² + b²)

Answer 2

Let u = sin(ax) and dv = sin(bx)dx

Then, du = acos(ax)dx and v = -(1/b)cos(bx)

So, ∫sin(ax)sin(bx)dx = uv - ∫vdu = -(1/b)sin(ax)cos(bx) + (a/b) ∫cos(ax)cos(bx)dx

Now let u = cos(ax) and dv = cos(bx)dx

So, du = -asin(ax) dx and v = (1/b)sin(bx)

Thus, ∫sin(ax)sin(bx)dx = -(1/b)sin(ax)cos(bx) + (a/b)[(1/b)cos(ax)sin(bx) + (a/b)∫sin(ax)sin(bx)dx]

Combining like terms, we get (1-a²/b²)∫sin(ax)sin(bx)dx = -(1/b)sin(ax)cos(bx) + a/b²cos(ax)sin(bx)

Then, divide both sides by 1 - a²/b²

Answer 3

Let u = cos(ax) and dv = cos(bx)dx. Then du = -asin(ax)dx and v = sin(bx)/b. We have:

∫ cos(ax)cos(bx)dx = cos(ax)sin(bx)/b + (a/b) ∫sin(ax)sin(bx)dx

Now, let u = sin(ax) and dv = sin(bx)dx. Then du = acos(ax) and v = -cos(bx)/b. We have:

∫ cos(ax)cos(bx)dx = cos(ax)sin(bx)/b - asin(ax)cos(bx)/b² + (a²/b²) ∫ cos(ax)cos(bx)dx

Now, we want to solve the last equation for ∫ cos(ax)cos(bx)dx. We have:

[1-(a²/b²)] ∫ cos(ax)cos(bx)dx = cos(ax)sin(bx)/b - asin(ax)cos(bx)/b² = (bcos(ax)sin(bx)- asin(ax)cos(bx))/b²

∫ cos(ax)cos(bx)dx = (bcos(ax)sin(bx)- asin(ax)cos(bx))/[b²(1-(a²/b²))]

∫ cos(ax)cos(bx)dx = [asin(ax)cos(bx) - bcos(ax)sin(bx)]/(a²-b²)

Answer 4

Let u = cos(ax) and dv = sin(bx)dx. Then du = -asin(ax)dx and v = -cos(bx)/b. We have:

∫ cos(ax)sin(bx)dx = -cos(ax)cos(bx)/b - (a/b) ∫sin(ax)cos(bx)dx

Now let u = sin(ax) and dv = cos(bx)dx. Then, du = acos(ax)dx and v = sin(bx)/b. We have:

∫ cos(ax)sin(bx)dx = -cos(ax)cos(bx)/b - asin(ax)sin(bx)/b²+ (a²/b²) ∫ cos(ax)sin(bx)dx

Now, we want to solve the last equation for ∫ cos(ax)sin(bx)dx. We have:

(1-(a²/b²)) ∫ cos(ax)sin(bx)dx = -cos(ax)cos(bx)/b - asin(ax)sin(bx)/b²

(1-(a²/b²)) ∫ cos(ax)sin(bx)dx = -(bcos(ax)cos(bx) +asin(ax)sin(bx))/b²

∫ cos(ax)sin(bx)dx = (bcos(ax)cos(bx) +asin(ax)sin(bx))/(((a²/b²) - 1)b²)

∫ cos(ax)sin(bx)dx = (bcos(ax)cos(bx) +asin(ax)sin(bx))/(a² - b²)

Answer 5

Find ∫ e^x•cos(x) dx

∫ u dv = uv - ∫ v du

Let u = e^x dv = cos(x) dx

du = e^x dx v = sin(x)

∫ e^x cos(x) dx = e^x •sin(x) - ∫ e^x sin(x) dx (1)

Looking at ∫ e^x sin(x) dx, apply integration by parts again:

Let w = e^x dz = sin(x) dx

dw = e^x dx z = -cos(x)

∫ w dz = wz - ∫ z dw

∫ e^x sin(x) dx = -e^x cos(x) - ∫ -cos(x) e^x dx

Now (1) can be re-written:

∫ e^x cos(x) dx = e^x sin(x) - (-e^x cos(x) - ∫ -cos(x) e^x dx )

Being careful with the signs, remove the paren's:

∫ e^x cos(x) dx = e^x sin(x) + e^x cos(x) - ∫ cos(x) e^x dx

Noting that the last term subtracted is the original integral (!), bring it over to the LHS:

2 ∫ e^x cos(x) dx = e^x ( sin(x) + cos(x) )

Finally, we get:

∫ e^x cos(x) dx = (1/2) e^x ( sin(x) + cos(x) ) + C

where C is some constant of integration, for the indefinite integral case.

Answer 6

The inspiration for integration by parts comes from the Product Rule when differentiating. Referring to f(x) and g(x) as f and g, respectively, for less cluttering:

d/dx(f * g) = f * g´ + g * f´

Integrating both sides we get

f * g = ∫(f * g´)dx + ∫(g * f´)dx, therefore

f * g - ∫(f * g´)dx = ∫(g * f´)dx <~~ equation (1) ~~

Since we haven't defined the functions f and g explicitly, there's some symmetry in this expression. So all that matters is that the integral we have to evaluate will change from ∫(g * f´) to ∫(f * g´). With this in mind, we can see that in the given example the two functions are e^x and cos(x).

In this particular example, it just so happens that it doesn't matter which function we choose for g and f, we can reach the answer in a similar way (good exercise to try out), so we'll choose f = cos(x) and g = e^x. Then, for the right hand side on expression (1) above,

∫(g * f´)dx = ∫(e^x) * (cos(x)) dx,

and now can calculate the terms on the left hand side,

g * f = e^x * sin(x), and

∫(f * g´)dx = ∫(e^x) * (sin(x)) dx,

=> e^x * sin(x) - ∫(e^x) * (sin(x)) dx = ∫(e^x) * (cos(x)) dx <~~ equation (2) ~~

From here, we do integration by parts again for the integral on the left hand side, ∫(e^x) * (sin(x)) dx. If we choose e^x as f and sin(x) as g using equation (1),

e^x * sin(x) - e^x * sin(x) + ∫(e^x) * (cos(x)) dx = ∫(e^x) * (cos(x)) dx,

=> 0 = 0.

So instead we should choose e^x to differentiate and sin(x) to integrate, namely e^x as g and sin(x) as f in equation (1), and

-e^(x) * cos(x) + ∫(e^x) * (cos(x)) dx = ∫(e^x) * (sin(x)) dx.

Then we substitute ∫(e^x) * (sin(x)) dx in equation (2) for the left hand side above, and we get

e^x * sin(x) + e^(x) * cos(x) - ∫(e^x) * (cos(x)) dx = ∫(e^x) * (cos(x)) dx,

e^x * (sin(x) + cos(x)) = 2 * ∫(e^x) * (cos(x)) dx,

e^x * (sin(x) + cos(x)) / 2 = ∫(e^x) * (cos(x)) dx,

where the value in the left hand side equals ∫(e^x) * (cos(x)) dx. So we calculated the expression for the original integral using integration by parts twice. Hope that helps!