What is the integral of (x^2+1)sin(2x-3)dx ?

The integral of (x^2+1)sin(2x-3)dx is 1/4 (-2x^2cos(2x-3)+2xsin(2x-3)-cos(2x-3))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2+1)sin(2x-3)dx

integral

(x^{2} + 1) sin (2 x - 3) d x

\frac{1}{4} (- 2 x^{2} cos (2 x - 3) + 2 x sin (2 x - 3) - cos (2 x - 3)) + C

Solution steps

\int (x^{2} + 1) sin (2 x - 3) d x

Apply u-substitution

: \int \frac{sin ( u ) ( u ^{2} + 6 u + 13 )}{8} d u

= \int \frac{sin ( u ) ( u ^{2} + 6 u + 13 )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sin (u) (u^{2} + 6 u + 13) d u

Expand

sin (u) (u^{2} + 6 u + 13) : u^{2} sin (u) + 6 u sin (u) + 13 sin (u)

= \frac{1}{8} \cdot \int u^{2} sin (u) + 6 u sin (u) + 13 sin (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int u^{2} sin (u) d u + \int 6 u sin (u) d u + \int 13 sin (u) d u)

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

\int 6 u sin (u) d u = 6 (- u cos (u) + sin (u))

\int 13 sin (u) d u = - 13 cos (u)

= \frac{1}{8} (- u^{2} cos (u) + 2 (u sin (u) + cos (u)) + 6 (- u cos (u) + sin (u)) - 13 cos (u))

Substitute back

u = 2 x - 3

= \frac{1}{8} (- (2 x - 3)^{2} cos (2 x - 3) + 2 ((2 x - 3) sin (2 x - 3) + cos (2 x - 3)) + 6 (- (2 x - 3) cos (2 x - 3) + sin (2 x - 3)) - 13 cos (2 x - 3))

Simplify

\frac{1}{8} (- (2 x - 3)^{2} cos (2 x - 3) + 2 ((2 x - 3) sin (2 x - 3) + cos (2 x - 3)) + 6 (- (2 x - 3) cos (2 x - 3) + sin (2 x - 3)) - 13 cos (2 x - 3)) : \frac{1}{4} (- 2 x^{2} cos (2 x - 3) + 2 x sin (2 x - 3) - cos (2 x - 3))

= \frac{1}{4} (- 2 x^{2} cos (2 x - 3) + 2 x sin (2 x - 3) - cos (2 x - 3))

Add a constant to the solution

= \frac{1}{4} (- 2 x^{2} cos (2 x - 3) + 2 x sin (2 x - 3) - cos (2 x - 3)) + C

\int \frac{2}{5 x} - \frac{3}{x ^{2}} d x

\int \frac{6 x ^{2}}{x ^{3} - 2} d x

\int \frac{2 x ^{2} - 3 x + 2}{x ^{3} + x} d x

\int 2^{t a n (x)} sec^{2} (x) d x

\int 4 x + 50 e^{- 0.2 x} d x

What is the integral of (x^2+1)sin(2x-3)dx ?
The integral of (x^2+1)sin(2x-3)dx is 1/4 (-2x^2cos(2x-3)+2xsin(2x-3)-cos(2x-3))+C