Question

Evaluvate intagiral (x³+2x²+3)sinx dx

Answer 1

Evaluate :-

$\rm \: \int \: ( {x}^{3} + {2x}^{2} + 3)sinx \: dx$

─━─━─━─━─━─━─━─━─━─━─

Method used :-

Integration by Parts

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

✏️See the rule:

∫u v dx = u∫v dx −∫u' (∫v dx) dx

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

• I - Inverse trigonometric functions

• L -Logarithmic functions

• A - Arithmetic and Algebraic functions

• T - Trigonometric functions

• E- Exponential functions

The alphabet which comes first is choosen as u and other as v.

$\boxed{ \red{ \rm \: \int \:sinx \: dx \: = - cosx + c}}$

$\boxed{ \red{ \rm \: \int \:cosx \: dx \: = sinx + c}}$

$\boxed{ \red{ \rm \: \dfrac{d}{dx} \: {x}^{n} \: = n {x}^{n - 1} }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

$\rm : \implies \:Let \: I \: = \rm \: \int \: ( {x}^{3} + {2x}^{2} + 3)sinx \: dx$

Using formula of Integration By Parts, we get

$\rm :I = ({x}^{3} + {2x}^{2} + 3) \int \: sinxdx \: - \int \: (\dfrac{d}{dx} ({x}^{3} + {2x}^{2} + 3) \int \: sinxdx)$

$\rm :I = - ({x}^{3} + {2x}^{2} + 3)cosx + \int( {3x}^{2} + 4x)cosxdx$

$\rm :I = - ({x}^{3} + {2x}^{2} + 3)cos \: + \: I_1 - - - (i)$

where,

$\rm :I_1 = \int(3{x}^{2} + {4x})cosxdx$

Using formula of Integration By Parts, we get

$\rm :I = (3{x}^{2} + {4x}) \int \: cosxdx \: - \int \: ( \dfrac{d}{dx} (3{x}^{2} + {4x}) \int \: cosxdx)dx$

$\rm :I_1 = (3{x}^{2} + {4x})sinx - \int \: (6x + 4)sinxdx$

$\rm :I_1 = (3{x}^{2} + {4x})sinx - I_2 - - (ii)$

where,

$\rm :I_2 = \int \: (6x + 4) \: sinx \: dx$

Using formula of Integration By Parts, we get

$\rm :I_2 = (6x + 4)\int \:sinxdx - \int \:(\dfrac{d}{dx} (6x + 4)\int \:sinxdx)dx$

$\rm :I_2 = - (6x + 4)cosx + \int \:6cosxdx$

$\rm :I_2 = - (6x + 4)cosx \: + \: 6sinx \: + \: c - -(iii)$

Put value from equation (iii) in equation (ii), we get

$\rm :I_1 = (3{x}^{2} + {4x})sinx + (6x + 4)cosx - 6sinx - c$

$\rm :I_1 = (3{x}^{2} + {4x} - 6)sinx \: + (6x + 4)cosx - c - - (iv)$

Now, Put value of equation (iv) in equation (i), we get

$\rm :I = - ({x}^{3} + {2x}^{2} + 3)cos \: + \: (3{x}^{2} + {4x} - 6)sinx \: + (6x + 4)cosx - c$

$\rm :I = - ({x}^{3} + {2x}^{2} + 3 - 6x - 4)cos \: + (3{x}^{2} + {4x} - 6)sinx \: - c$

$\rm :I = - ({x}^{3} + {2x}^{2} - 6x - 1)cos \: + (3{x}^{2} + {4x} - 6)sinx \: - c$

Answer 2

We're given to evaluate,

$\displaystyle\longrightarrow I=\int\left(x^3+2x^2+3\right)\sin x\ dx$

$\displaystyle\longrightarrow I=\int\left(x^3\sin x+2x^2\sin x+3\sin x \right)\ dx$

$\displaystyle\longrightarrow I=\int x^3\sin x\ dx+2\int x^2\sin x\ dx+3\int\sin x\ dx\quad\quad\dots(1)$

Consider,

$\displaystyle\longrightarrow I_1=\int x^3\sin x\ dx$

Instead of doing integration by parts directly, here we can make use of the following if $f$ is an algebraic function of integral power and $g$ is trigonometric or exponential function.

$\displaystyle\longrightarrow\int fg\ dx=fg_{-1}-f_1g_{-2}+f_2g_{-3}-f_3g_{-4}+f_4g_{-5}-\,\dots\,(+C)$

until we get $f_k=0$ for some $k,$ where,

• $f_n$ is $n^{th}$ derivative of $f$

• $g_{-n}$ is $n^{th}$ integral of $g$

So,

$\displaystyle\longrightarrow I_1=\int x^3\sin x\ dx$

$\displaystyle\longrightarrow I_1=x^3(-\cos x)-3x^2(-\sin x)+6x(\cos x)-6(\sin x)+0(-\cos x)+C_1$

$\displaystyle\longrightarrow I_1=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C_1$

Similarly, consider,

$\displaystyle\longrightarrow I_2=\int x^2\sin x\ dx$

$\displaystyle\longrightarrow I_2=x^2(-\cos x)-2x(-\sin x)+2(\cos x)-0(\sin x)+C_2$

$\displaystyle\longrightarrow I_2=-x^2\cos x+2x\sin x+2\cos x+C_2$

Consider,

$\displaystyle\longrightarrow I_3=\int\sin x\ dx$

$\displaystyle\longrightarrow I_3=-\cos x+C_3$

Then (1) becomes,

$\displaystyle\begin{aligned}\longrightarrow I=\ \ &\left(-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C_1\right)\\+\ \ &2\left(-x^2\cos x+2x\sin x+2\cos x+C_2\right)\\+\ \ &3\left(-\cos x+C_3\right)\end{aligned}$

$\displaystyle\longrightarrow I=\left(3x^2+4x-6\right)\sin x-\left(x^3+2x^2-6x-1\right)\cos x+\left(C_1+2C_2+3C_3\right)$

Taking $C_1+2C_2+3C_3=C,$

$\displaystyle\longrightarrow\underline{\underline{I=\left(3x^2+4x-6\right)\sin x-\left(x^3+2x^2-6x-1\right)\cos x+C}}$