Substitution and Integration by Parts — Basic Exercises

date2026-07-15document_iddoc_5a2fcb57d7d68755210bf841cb2c3e7edescription置換積分と部分積分を、連鎖律と積の微分公式の逆として確認する基本演習である。prerequisites積分公式と計算法 / 微分公式と計算法type問題演習content_typeexercisestatusactive

mathcalculusexercisesubstitutionintegration-by-parts

data/lecture/math/calculus/integration-rules-and-computation.lecture.n.md

1Problem 1

Evaluate

\int 2 x \sin (x^{2}) d x

and verify the answer by differentiation.

1.1Sample Solution

Let $f (u) = \sin u$ and $g (x) = x^{2}$ . The integrand then has the form $f (g (x)) g^{'} (x)$ , with $g^{'} (x) = 2 x$ . Let $u = x^{2}$ , so $d u = 2 x d x$ . Then

\int 2 x \sin (x^{2}) d x = \int \sin u d u = - \cos u + C = - \cos (x^{2}) + C .

Moreover,

\frac{d}{d x} (- \cos (x^{2}) + C) = 2 x \sin (x^{2}),

which recovers the original integrand.

1.2Explanation

Substitution reverses the chain rule. Check that the integrand contains both a composite function $f (g (x))$ and the derivative $g^{'} (x)$ of its inner function as factors.

2Problem 2

Evaluate

\int x \cos x d x

and verify the answer by differentiation.

2.1Sample Solution

Choose $u = x$ and $v^{'} = \cos x$ , so $u^{'} = 1$ and $v = \sin x$ . Integration by parts gives

\int x \cos x d x = x \sin x - \int \sin x d x = x \sin x + \cos x + C .

Furthermore,

\frac{d}{d x} (x \sin x + \cos x + C) = \sin x + x \cos x - \sin x = x \cos x .

2.2Explanation

Integration by parts reverses the product rule. Differentiating the polynomial factor $x$ simplifies it to $1$ , while an antiderivative of $\cos x$ is readily available.

3Comprehensive Exercises

data/exercise/math/calculus/integration-methods-and-computation.exercise.n.md

4Related Lectures

data/lecture/math/calculus/differentiation-rules-and-computation.lecture.n.md data/lecture/math/calculus/integration-rules-and-computation.lecture.n.md