What is the integral from 0 to pi of t^2cos(nt) ?

The integral from 0 to pi of t^2cos(nt) is (2pi(-1)^n)/(n^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi of t^2cos(nt)

\int_{0}^{π} t^{2} cos (n t) d t

\frac{2 π ( - 1 ) ^{n}}{n ^{2}}

Solution steps

\int_{0}^{π} t^{2} cos (n t) d t

Apply u-substitution

: \int_{0}^{nπ} \frac{u ^{2} cos ( u )}{n ^{3}} d u

= \int_{0}^{nπ} \frac{u ^{2} cos ( u )}{n ^{3}} d u

Take the constant out:

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

= \frac{1}{n ^{3}} \cdot \int_{0}^{nπ} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{nπ}

= \frac{1}{n ^{3}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{nπ}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{n ^{3}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{nπ}

Compute the boundaries:

2 π (- 1)^{n} n

= \frac{1}{n ^{3}} \cdot 2 π (- 1)^{n} n

Simplify

= \frac{2 π ( - 1 ) ^{n}}{n ^{2}}

\int_{1}^{3} \frac{1}{( x - 1 ) ^{\frac{4}{3}}} d x

\int_{1}^{2} 2 x (x^{2} + 3) d x

\int_{y}^{1} x^{2} e^{x y} d x

\int_{- 1}^{0} 6 x^{2} (2 x^{3} + 5)^{2} d x

\int_{0}^{\infty} k e^{- 3 x} d x

What is the integral from 0 to pi of t^2cos(nt) ?
The integral from 0 to pi of t^2cos(nt) is (2pi(-1)^n)/(n^2)