integral from 0 to pi of 2xcos(nx)

해법

\int_{0}^{π} 2 x cos (n x) d x

\frac{2 ( ( - 1 ) ^{n} - 1 )}{n ^{2}}

솔루션 단계

\int_{0}^{π} 2 x cos (n x) d x

정수를 빼라:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{π} x cos (n x) d x

대체 적용

: \int_{0}^{nπ} \frac{u cos ( u )}{n ^{2}} d u

= 2 \cdot \int_{0}^{nπ} \frac{u cos ( u )}{n ^{2}} d u

정수를 빼라:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{n ^{2}} \cdot \int_{0}^{nπ} u cos (u) d u

부품별 통합 적용

: [u sin (u) - \int sin (u) d u]_{0}^{πn}

= 2 \cdot \frac{1}{n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{πn}

\int sin (u) d u = - cos (u)

= 2 \cdot \frac{1}{n ^{2}} [u sin (u) - (- cos (u))]_{0}^{πn}

2 \cdot \frac{1}{n ^{2}} [u sin (u) - (- cos (u))]_{0}^{πn}

간소화하다 :

\frac{2}{n ^{2}} [u sin (u) + cos (u)]_{0}^{πn}

= \frac{2}{n ^{2}} [u sin (u) + cos (u)]_{0}^{πn}

경계 계산:

(- 1)^{n} - 1

= \frac{2}{n ^{2}} ((- 1)^{n} - 1)

단순화

= \frac{2 ( ( - 1 ) ^{n} - 1 )}{n ^{2}}