Mathematics

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

##### SOLUTION
$\int { { x }^{ 2 }\sin2xdx } \\ ={ x }^{ 2 }\int { \sin2xdx } -\int { \left[ \dfrac { d }{ dx } { x }^{ 2 }.\int { \sin2xdx } \right] } dx\\ =-\dfrac { { x }^{ 2 } }{ 2 } \cos2x+\int { x.\cos2xdx } \\ =-\dfrac { { x }^{ 2 } }{ 2 } \cos2x+\dfrac { x }{ 2 } \sin2x-\dfrac { \cos2x }{ 4 } +C\\ =\dfrac { 2x\sin2x-2{ x }^{ 2 }\cos2x-\cos2x }{ 4 } +C$

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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 Hard
If $x$ satisfies the equation $\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$
for $(0<\alpha<\pi)$
then the value of $x$ is?
• A. $\displaystyle\pm\sqrt{\frac{\alpha}{2\sin{\alpha}}}$
• B. $\displaystyle\pm\sqrt{\frac{2\sin{\alpha}}{\alpha}}$
• C. $\displaystyle\pm\sqrt{\frac{\alpha}{\sin{\alpha}}}$
• D. $\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
$\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( x+\sqrt{1+x^{2}} \right )^{n}} = f(n)$
Find the value if $n=6$, can be expressed as $a/b$ in simplest form, then $b-a = ?$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Integrate $\int {({{\sin }^{ - 1}}} x{)^2}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
The value of $\int\dfrac{x^2+1}{(x+1)^2}dx$  is

Evaluate $\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$