Mathematics

# Solve:$\int x^2e^x\sin xdx$

##### SOLUTION
$I=\displaystyle\int x^{2}e^{x}\sin xdx$
Integrating by parts $u=x^{2}. V=e^{x}\sin x$
$u^{1}=2x, \displaystyle\int vdx=\dfrac{1}{2}e^{x} (\sin x-\cos x)$  [Integrated by parts]
$I=x^{2}\displaystyle\int e^{x}\sin dx-\int 2x \left(\int e^{x}\sin x dx\right)dx$
$=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)-\int xe^{x}(\sin x-\cos x)dx$
Now, $\displaystyle\int e^{x}(\cos x-\sin x)dx=e^{x}\cos x [\because \int e^{x}f(x)+f'(x)dx=e^{x}f(x)]$
$I=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)+\left[x e^{x}\cos x-\int (1) e^{x}\cos xdx\right]$     [Using by parts]
$=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)+xe^{x}\cos x-\dfrac{e^{x}}{2} (\sin x+\cos x)+C$ [Using by parts]
$=\dfrac{e^{x}}{2}[\sin (x)(x^{2}-1)+\cos x(-x^{2}+2x-1)]+C$
$I=\dfrac{e^{x}}{2}((x^{2}-1)\sin x-(x-1)^{2}\cos x)+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
Evaluate $\displaystyle\int \frac{x^{2}}{9 + 16 x^{6}} dx$
• A. $\displaystyle \frac{1}{16} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$
• B. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{3x^{3}}{4} \right) + c$
• C. $\displaystyle \frac{1}{16} \tan^{-1} \left (\displaystyle \frac{3x^{3}}{4} \right) + c$
• D. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate: $\displaystyle \int \dfrac{1}{1+tanx}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\frac { d } { d x } f ( x ) = g ( x )$ then the value of $\int _ { a } ^ { b } f ( x ) g ( x ) d x$ is
• A. $f ( b ) - f ( a )$
• B. $g ( b ) - g ( a )$
• C. $\frac { 1 } { 2 } \left[ \{ f ( b ) \} ^ { 2 } - \{ f ( a ) \} ^ { 2 } \right]$
• D. $\frac { 1 } { 2 } \left[ \{ g ( b ) \} ^ { 2 } - \{ g ( a ) \} ^ { 2 } \right]$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The integral $\displaystyle\int^{{\pi}/{4}}_{{\pi}/{12}}\frac{8\cos 2x}{(\tan x+\cot x)^3}dx$ equals?
• A. $\displaystyle\frac{13}{32}$
• B. $\displaystyle\frac{13}{256}$
• C. $\displaystyle\frac{15}{64}$
• D. $\displaystyle\frac{15}{128}$

Given that for each $\displaystyle a \in (0, 1), \lim_{h \rightarrow 0^+} \int_h^{1-h} t^{-a} (1 -t)^{a-1}dt$ exists. Let this limit be $g(a)$
In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$