Match List – I with List – II.
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LIST I |
LIST II |
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A. $\int\sin ^2(2 x+1) d x$ |
I. $\frac{x}{2}+\frac{\sin (2 x+1) \cos (2 x+1)}{4} + C$ : C is a constant |
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B. $\int\cos ^2(2 x+1) d x$ |
II. $\frac{\cos 3(2 x+1)}{24}-\frac{3 \cos (2 x+1)}{8}+C$ : C is a constant |
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C. $\int\sin ^3(2 x+1) d x$ |
III. $\frac{x}{2}-\frac{\sin (2 x+1) \cos (2 x+1)}{4}+C$ : C is a constant |
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D. $\int\cos ^3(2 x+1) d x$ |
IV. $\frac{\sin 3(2 x+1)}{24}+\frac{3 \sin (2 x+1)}{8}+C$ : C is a constant |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - A-III, B-I, C-II, D-IV
(A) $\int\sin ^2(2 x+1) d x$
$=\frac{\int 1-\cos(4x+2)dx}{2}=\frac{x}{2}-\frac{\sin(4x+2)}{2×4}+C$
$=\frac{x}{2}-\frac{\sin(2x+1)\cos(2x+1)}{4}+C$ (III)
(B) $\int\cos ^2(2 x+1) d x$
$=\frac{\int 1+\cos(4x+2)dx}{2}=\frac{x}{2}+\frac{\sin(4x+2)}{2×4}+C$
$=\frac{x}{2}+\frac{\sin(2x+1)\cos(2x+1)}{4}+C$ (I)
(C) $\int\sin^3(2 x+1) d x$
$\int\sin^3(x)dx=\int\frac{3\sin x-\sin 3x}{4}dx$
so $\int\sin^3(2 x+1)=\int\frac{3\sin (2 x+1)-\sin (3(2 x+1))}{4}dx$
$=\frac{\cos 3(2 x+1)}{24}-\frac{3 \cos (2 x+1)}{8}+C$ (II)
(D) $\int\cos ^3(2 x+1) d x$
$=\int\frac{\cos 3(2 x+1)+3\cos (2 x+1)}{4}dx$
$\frac{\sin 3(2 x+1)}{24}+\frac{3 \sin (2 x+1)}{8}+C$ (IV)