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Home Trigonometric Functions Integration by Parts Examples Example 3: Indefinite Integration by Parts
 
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Example 3: Indefinite Integration by Parts

Evaluate ∫ x2 sin x dx. This requires two integrations by parts.

Step 1

u = x2, dv = sin x dx,

du = 2x dx, ∫ dv = ∫ sin x dx = - cos x + C.

We take v = - cos x.

∫ x2 sin x dx = uv - ∫ v du = -x2 cos x + ∫ 2x cos x dx.
Step 2

Evaluate ∫ 2x cos x dx.

u1 = 2x,             dv1 = cos x dx,

du1 = 2 dx,          ∫ dv1 = ∫ cos x dx = sin x + C.

We take v1 = sin x.

∫ 2x cos x dx = u1v1 - ∫ v1 du1

= 2x sin x - ∫ 2 sin x dx

= 2x sin x + 2 cos x + C.

Combining the two steps,

∫ x2 sin x dx = -x2 cos x + 2x sin x + 2 cos x + C.

Sometimes integration by parts will yield an equation in which the given integral occurs on both sides. One can often solve for the answer.
Home Trigonometric Functions Integration by Parts Examples Example 3: Indefinite Integration by Parts

Last Update: 2006-11-15