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