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 
$$\displaystyle \frac{1}{2}\left [ x^{3}+\frac{x^{2}}{2}-2x \right ]$$
$$\displaystyle -\frac{1}{2}\left [ \left ( 3x^{2}+x-2 \right )\frac{\sin \left ( 6x+2 \right )}{6}+\left ( 6x+1 \right )\frac{\cos\left ( 6x+2 \right )}{36}-\frac{\sin \left ( 6x+2 \right )}{6\times 36} \right ]+C$$
$$\displaystyle -\frac{1}{2}\left [ x^{3}+\frac{x^{2}}{2}-2x \right ]-\frac{18x^{2}+6x-13}{72}\sin \left ( 6x+2 \right )-\frac{1}{72}\left ( 6x+1 \right )\cos\left ( 6x+2 \right )+C.$$
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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


ANSWER

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


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


ANSWER

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


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
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