#### Passage

In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts.

$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $\int P_{n}(x)\, Q(x)\, dx$, where $P_{n}(x)$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.

Mathematics

# If $\int e^{2x}. x^{4}\, dx\, =\, \displaystyle \frac{e^{2x}}{2} f(x)\, +\, C$ then f(x) is equal to

$x^{4}\, -\, 2x^{3}\, +\, 3x^{2}\, -\, 3x\, +\, \displaystyle \frac{3}{2}$

##### SOLUTION
$I=\int e^{2x}. x^{4} dx$
Applying integration by parts,
$\displaystyle = x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -\int { { 4x }^{ 3 } } \frac { e^{ 2x } }{ 2 } dx$

$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -2\int { { x }^{ 3 } } e^{ 2x }dx$
Again applying integration by parts
$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -2{ x }^{ 3 }\frac { e^{ 2x } }{ 2 } +2\int { { 3x }^{ 2 } } \frac { e^{ 2x } }{ 2 } dx$

$\Rightarrow \displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+3 \int { { x }^{ 2 } } e^{ 2x }dx$

Again applying integration by parts, we get
$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+3x^{ 2 }\, \frac { e^{ 2x } }{ 2 } -3\int { { 2x } } \frac { e^{ 2x } }{ 2 } dx$

$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+\frac { 3 }{ 2 } x^{ 2 }\, e^{ 2x }-3 \int { { x } } e^{ 2x }dx$

Again applying integration by parts, we get
$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+\frac { 3 }{ 2 } x^{ 2 }\, e^{ 2x }-\frac { 3 }{ 2 } xe^{ 2x }+3 \int { \frac { e^{ 2x } }{ 2 } } dx$

$\Rightarrow \displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+\frac { 3 }{ 2 } x^{ 2 }\, e^{ 2x }-\frac { 3 }{ 2 } xe^{ 2x }+\frac { 3 }{ 4 } e^{ 2x }+C$

$\Rightarrow \displaystyle I=\dfrac{e^{2x}}{2} (x^4-2x^3+3x^2-3x+\frac{3}{2})+C$

On comparing with given , we get
$f(x)=x^4-2x^3+3x^2-3x+\dfrac{3}{2}$

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Single Correct Hard Published on 17th 09, 2020
Mathematics

# If $\int (x^{3}\, -\, 2x^{2}\, +\, 3x\, -\, 1)\, cos2x\, dx\, =\, \displaystyle \frac{sin 2x}{4}u(x)\, +\, \frac{cos 2x}{8}v(x)\, +\, c$, then

$u(x)\, =\, 2x^{3}\, -\, 4x^{2}\, +\, 3x$

##### SOLUTION
$\int { \left( { x }^{ 2 }-2{ x }^{ 2 }+3x-1 \right) } \cos { 2x } dx\\ =\int { { x }^{ 2 } } \cos { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } } dx+\int { 3x\cos { 2x } } dx-\int { \cos { 2x } } dx\\ =\frac { 1 }{ 2 } { x }^{ 3 }\sin { 2x } -\frac { 3 }{ 2 } \int { { x }^{ 2 } } \sin { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } } dx+3\int { x } \cos { 2x } d-\int { \cos { 2x } dx }$
$\displaystyle =\frac { 3 }{ 4 } { x }^{ 2 }\cos { 2x } +\frac { 1 }{ 2 } { x }^{ 3 }\sin { 2x } +\frac { 3 }{ 2 } \int { x } \cos { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } dx } -\int { \cos { 2x } } dx$
$\displaystyle ={ x }^{ 3 }\sin { x } \cos { x- } { x }^{ 2 }\sin { 2x } +\frac { 3 }{ 4 } { x }^{ 2 }\cos { 2x } \frac { 3 }{ 4 } x\sin { 2x } -x\cos { 2x } +\frac { 3 }{ 8 } \cos { 2x }$
$\displaystyle =\frac { 1 }{ 4 } \left( { 2x }^{ 3 }-{ 4x }^{ 2 }+3x \right) \sin { 2x } +\frac { 1 }{ 8 } \left( { 12x }^{ 3 }-{ 16x }^{ 2 }+6x \right) \cos { 2x }$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Multiple Correct Hard
Let $f(x)\, =\, 3x^{2}.\, \sin \,\displaystyle \frac{1}{x}\, -\, x\cos\, \displaystyle \frac{1}{x},\, x\, \neq\, 0, f(0)\, =\, 0\, f \left (\, \displaystyle \frac{1}{\pi} \right )\, =\, 0$, then which of the  following is/are not correct.
• A. $f(x)$ is continuous at $x = 0$
• B. $f(x)$ is non-differentiable at $x = 0$
• C. $f(x)$ is discontinuous at $x = 0$
• D. $f(x)$ is differentiable at $x = 0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
solve it.
$I = \int {{x^2}\cos x} dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\displaystyle\int^a_0\dfrac{dx}{(ax+a^2-x^2)}$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate:$\displaystyle\int_{0}^{\frac{\pi}{2}}{{\sin}^{3}{x}\,dx}$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$