#### 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
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
where  $\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$
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 $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.
Mathematics

# If $\displaystyle \int \left ( 3x^{2}+x-2 \right )\sin ^{2}\left ( 3x+1 \right )dx$$\displaystyle =-\frac{u\left ( x \right )}{72}\sin \left ( 6x+2 \right )+\frac{v\left ( x \right )}{72}\cos \left ( 6x+2 \right )+\frac{1}{2}x^{3}+\frac{1}{4}x^{2}-x+C ##### ANSWER \displaystyle v\left ( x \right )=3-\left ( 6x+1 \right ) ##### SOLUTION Let I=\displaystyle \int \left ( 3x^{2}+x-2 \right )\sin ^{2}\left ( 3x+1 \right )dx = \displaystyle \frac{1}{2}\int \left ( 3x^{2}+x-2 \right )\left ( 1-\cos \left ( 6x+2 \right ) \right ). Now applying the formula in comprehension, the last integral can be written as \displaystyle \frac{1}{2}\left [ x^{3}+\frac{x^{2}}{2}-2x \right ] \displaystyle -\frac{1}{2}\left [ \left ( 3x^{2}+x-2 \right )\frac{\sin \left ( 6x+2 \right )}{6}+\left ( 6x+1 \right )\frac{\cos\left ( 6x+2 \right )}{36}-\frac{\sin \left ( 6x+2 \right )}{6\times 36} \right ]+C \displaystyle -\frac{1}{2}\left [ x^{3}+\frac{x^{2}}{2}-2x \right ]-\frac{18x^{2}+6x-13}{72}\sin \left ( 6x+2 \right )-\frac{1}{72}\left ( 6x+1 \right )\cos\left ( 6x+2 \right )+C. Its FREE, you're just one step away Create your Digital Resume For FREE on your name's sub domain "yourname.wcard.io". Register Here! Single Correct Hard Published on 17th 09, 2020 Mathematics # If \displaystyle \int \left ( 2x^{3}+3x^{2}-8x+1 \right )\sqrt{2x+6}dx$$\displaystyle =\frac{\sqrt{2x+6}}{5.7.9}\left ( 2x+6 \right )f\left ( x \right )+C$ then f(x) is equal to

$\displaystyle 70x^{3}-45x^{2}-396x+897$

##### SOLUTION
applying the above formula in the comprehesion, the given integral is equal to
$=\displaystyle \left ( 2x^{3}+3x^{2}-8x+1 \right )\frac{\left ( 2x+6 \right )^{3/2}}{3}-\left ( 6x^{2}+6x-8 \right )\frac{\left ( 2x+6 \right )^{5/2}}{3.5}+\left ( 12x+6 \right )\frac{\left ( 2x+6 \right )^{7/2}}{3.5.7}-12\frac{\left ( 2x+6 \right )^{9/2}}{3.5.7.9}+C$
$\displaystyle =\frac{\sqrt{2x+6}}{5.7.9}\left ( 2x+6 \right )\left ( 70x^{3}-45x^{2}-396x+897 \right )+C$

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Single Correct Hard Published on 17th 09, 2020
Mathematics

# If $\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$$\displaystyle =\frac{\sin 2x}{4}u\left ( x \right )+\frac{\cos 2x}{8}v\left ( x \right )+C$ then

$\displaystyle u\left ( x \right )=2x^{3}-4x^{2}+3x$

##### SOLUTION
$\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$ $\displaystyle =\frac{\sin 2x}{4}u\left ( x \right )+\frac{\cos 2x}{8}v\left ( x \right )+C$ .... $(i)$

Applying the formula given in the comprehension, required integral becomes
$\displaystyle \int \left ( x^{3}-2x^{2}+3x-1 \right )\cos 2x\:dx$

$=\displaystyle \left (x^{3}-2x^{2}+3x-1 \right )\frac{\sin 2x}{2}-\left ( 3x^{2}-4x+3 \right )\left ( -\frac{\cos 2x}{4} \right )+\left ( 6x-4 \right )\left (-\dfrac{\sin 2x}{8} \right ) - 6\dfrac{\cos 2x}{16}+C$

$=\displaystyle \dfrac{\sin 2x}{4}(2x^{3}-4x^{2}+6x-2)-(3x-2)\dfrac{\sin 2x}{4}+\dfrac{\cos 2x}{8}(6x^{2}-8x+6)-3\dfrac{\cos 2x}{8}+C$

$\displaystyle =\frac{\sin 2x}{4}\left [ 2x^{3}-4x^{2}+3x \right ]+\frac{\cos 2x}{8}\left [ 6x^{2}-8x+3 \right ]+C$

Comparing with the given equation $(i)$, we get
$\displaystyle u\left ( x \right )=2x^{3}-4x^{2}+3x$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 128

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\int^{\pi /2}_{-\pi /2}\cos x$ ln $\left(\displaystyle\frac{1+x}{1-x}\right)dx$ is equal to.
• A. $\displaystyle\frac{\pi^2}{4}\left(-1+\frac{\pi}{2}\right)$
• B. $1$
• C. $\displaystyle\frac{\pi^2}{2}$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate using limit of sum:
$\displaystyle \int_{1}^{3} {(x+1)^2}dx$
• A. $26$
• B. $30$
• C. $32$
• D. $28$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Find the integral of    $\displaystyle \int (ax^2+bx+c)dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate :
$\int {\dfrac{1}{{\left( {x + 2} \right)\left( {x + 2} \right)}}} dx$

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