**Six Integrals from Calculus I: Substitution vs Integration by Parts**

Even slight differences integrands can change the strategy. Observe.

**Substitution**

For these three integrals, the substitution u = x^2 will serve well.

Observe:

u = x^2

du = 2 x dx

du / 2 = x dx

∫ x sin x^2 dx

= ∫ 1/2 sin u du

= -1/2 cos u + C

= -1/2 cos x^2 + C

∫ x cos x^2 dx

= ∫ 1/2 cos u du

= 1/2 sin u + C

= 1/2 sin x^2 + C

∫ x e^(x^2) dx

= ∫ 1/2 e^u du

= 1/2 e^u + C

= 1/2 e^(x^2) + C

**Integration by Parts**

General: ∫ u dv = u v - ∫ v du, v and u are functions (usually of x)

∫ x^2 sin x dx

u = x^2, dv = sin x dx; du = 2 x dx, v = - cos x

= -x^2 cos x + ∫ 2 x cos x dx

u = 2 x, dv = -cos x dx; du = 2 dx, v = -sin x

= -x^2 cos x + 2 x sin x - ∫ 2 sin x dx

= -x^2 cos x + 2 x sin x - ∫ 2 sin x dx

∫ x^2 cos x dx

u = x^2, dv = cos x dx; du = 2 x dx, v = sin x

= x^2 sin x - ∫ 2 x sin x dx

u = 2 x, dv = sin x dx; du = 2 dx, v = -cos x

= x^2 sin x + 2 x cos x - ∫ 2 cos x dx

= x^2 sin x + 2 x cos x - 2 sin x + C

∫ x^2 e^x dx

u = x^2, dv = e^x dx; du = 2 x dx, v = e^x

= x^2 e^x - ∫2 x e^x dx

u = 2 x, dv = e^x dx; du = 2 dx, v = e^x

= x^2 e^x - 2 x e^x + ∫2 e^x dx

= x^2 e^x - 2 x e^x + 2 e^x + C

= e^x (x^2 - 2 x + 2) + C

Bottom line, being good at integration, like everywhere else in math, will require practice, experience, and maybe later in life, some retraining.

Students, best of luck in your studies, in class and in life.

Eddie

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