Solutions — Integration by Parts — Grade 12 Specialist Maths

Find ∫xe^x dx. Fluency

By LIATE: Algebraic before Exponential, so let u = x and dv = e^x dx.

Then du = dx and v = e^x.

Applying ∫u dv = uv − ∫v du:

∫xe^x dx = xe^x − ∫e^x dx = xe^x − e^x + C

= e^x(x − 1) + C

Check by differentiating: d/dx[e^x(x−1)] = e^x(x−1) + e^x(1) = xe^x ✓

Find ∫x cos(x) dx. Fluency

By LIATE: Algebraic before Trigonometric. Let u = x, dv = cos x dx.

Then du = dx and v = sin x.

∫x cos x dx = x sin x − ∫sin x dx = x sin x − (−cos x) + C

= x sin x + cos x + C

Check: d/dx[x sin x + cos x] = sin x + x cos x − sin x = x cos x ✓

Find ∫ln(x) dx. Fluency

Write as ∫ln(x) × 1 dx. By LIATE: Logarithm first, so let u = ln x, dv = dx.

Then du = (1/x) dx and v = x.

∫ln x dx = x ln x − ∫x × (1/x) dx = x ln x − ∫1 dx = x ln x − x + C

= x(ln x − 1) + C

Check: d/dx[x ln x − x] = ln x + x × (1/x) − 1 = ln x + 1 − 1 = ln x ✓

Find ∫x²e^x dx. Fluency

The polynomial x² has degree 2, so two applications of by parts are needed.

First application: u = x², dv = e^x dx ⇒ du = 2x dx, v = e^x.

∫x²e^x dx = x²e^x − 2∫xe^x dx

Second application on ∫xe^x dx: u = x, dv = e^x dx ⇒ du = dx, v = e^x.

∫xe^x dx = xe^x − e^x

Combining: ∫x²e^x dx = x²e^x − 2(xe^x − e^x) + C

= x²e^x − 2xe^x + 2e^x + C

= e^x(x² − 2x + 2) + C

Find ∫x² sin(x) dx. Understanding

Requires two applications of by parts, keeping u as the polynomial each time.

First application: u = x², dv = sin x dx. ⇒ du = 2x dx, v = −cos x.

∫x²sin x dx = −x²cos x + 2∫x cos x dx

Second application on ∫x cos x dx: u = x, dv = cos x dx. ⇒ du = dx, v = sin x.

∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x

Combining: ∫x²sin x dx = −x²cos x + 2(x sin x + cos x) + C

= −x² cos x + 2x sin x + 2 cos x + C

Find ∫arctan(x) dx. Understanding

By LIATE: Inverse trig first. Write as ∫arctan(x) × 1 dx.

Let u = arctan x, dv = dx. ⇒ du = 1/(1+x²) dx, v = x.

∫arctan x dx = x arctan x − ∫ x/(1+x²) dx

For ∫ x/(1+x²) dx: Let w = 1+x², dw = 2x dx, so x dx = dw/2.

∫ x/(1+x²) dx = (1/2)∫ dw/w = (1/2) ln|w| = (1/2)ln(1+x²)

∫arctan x dx = x arctan x − (1/2)ln(1 + x²) + C

Find ∫e^xsin(x) dx using the cyclic method. Understanding

Let I = ∫e^xsin x dx.

First application: u = sin x, dv = e^x dx. ⇒ du = cos x dx, v = e^x.

I = e^xsin x − ∫e^xcos x dx

Second application on ∫e^xcos x dx: u = cos x, dv = e^x dx. ⇒ du = −sin x dx, v = e^x.

∫e^xcos x dx = e^xcos x − ∫e^x(−sin x) dx = e^xcos x + ∫e^xsin x dx = e^xcos x + I

Substituting: I = e^xsin x − (e^xcos x + I)

I = e^xsin x − e^xcos x − I

2I = e^x(sin x − cos x)

I = (e^x/2)(sin x − cos x) + C

Evaluate ∫₀¹ xe^−x dx. Understanding

Let u = x, dv = e^−x dx. ⇒ du = dx, v = −e^−x.

Applying the definite form ∫_a^b u dv = [uv]_a^b − ∫_a^b v du:

∫₀¹ xe^−x dx = [−xe^−x]₀¹ − ∫₀¹(−e^−x) dx

= [−xe^−x]₀¹ + ∫₀¹ e^−x dx

= [−xe^−x]₀¹ + [−e^−x]₀¹

= (−e⁻¹ − 0) + (−e⁻¹ − (−1))

= −e⁻¹ + 1 − e⁻¹

= 1 − 2e⁻¹

= 1 − 2/e ≈ 0.264

Find ∫(ln x)² dx. Problem Solving

Write as ∫(ln x)² × 1 dx. Let u = (ln x)², dv = dx.

du = 2 ln x × (1/x) dx = (2 ln x)/x dx, v = x.

∫(ln x)² dx = x(ln x)² − ∫x × (2 ln x)/x dx = x(ln x)² − 2∫ln x dx

From Q3: ∫ln x dx = x ln x − x.

∫(ln x)² dx = x(ln x)² − 2(x ln x − x) + C

= x(ln x)² − 2x ln x + 2x + C

= x[(ln x)² − 2 ln x + 2] + C

Check: d/dx{x[(ln x)²−2ln x+2]} = (ln x)²−2ln x+2 + x[2(ln x)/x − 2/x] = (ln x)²−2ln x+2+2ln x−2 = (ln x)² ✓

Find the volume of revolution of y = ln(x) from x = 1 to x = e, rotated about the x-axis. Problem Solving

The volume of revolution about the x-axis is V = π∫₁^e [ln(x)]² dx.

From Q9 (indefinite result): ∫(ln x)² dx = x[(ln x)² − 2 ln x + 2] + C.

Evaluate from x = 1 to x = e:

At x = e: e[(ln e)² − 2 ln e + 2] = e[1 − 2 + 2] = e[1] = e

At x = 1: 1[(ln 1)² − 2 ln 1 + 2] = 1[0 − 0 + 2] = 2

∫₁^e (ln x)² dx = e − 2

V = π(e − 2) ≈ 3.550 cubic units

Integration by Parts — Full Worked Solutions

Find ∫xex dx. Fluency