Match List – I with List – II.

Choose the correct answer from the options given below:

A-III, B-I, C-II, D-IV

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)