Match the following Column – I Column – II AIf ∫x14dx1+x=Ax14+Bx34+4tan−1x14+c, then A+BP1BIf ∫sinxdxsin(x−α)=Ax+Blnsin(x−α)+c, then A2+B2=Q−83CIf ∫dx(x2+1)(x2+4)=Atan−1x2+Btan−1x+c, then A+B=R14DIf ∫sin3xcos3x dx=k8[−32cos2x+16cos6x]+C then k=S16

A) x=t4⇒dx=4t3dt∫x14dx1+x=∫t4t3dt1+t2=4∫t4-11+t2+11+t2dt =4∫t2-1+11+t2dt=4t33-4t+4tan-1t+c B) ∫ sin((x−α)+α)sin(x−α)dx=cosα∫dx+sinα∫cot(x−α)dxC) ∫dx(x2+1)(x2+4)= 13∫1x2+1−1x2+4dxD) sin3xcos3x=18sin32x=18(3sin2x−sin6x4)

Match the following Column – I Column – II AIf ∫x14dx1+x=Ax14+Bx34+4tan−1x14+c, then A+BP1BIf ∫sinxdxsin(x−α)=Ax+Blnsin(x−α)+c, then A2+B2=Q−83CIf ∫dx(x2+1)(x2+4)=Atan−1x2+Btan−1x+c, then A+B=R14DIf ∫sin3xcos3x dx=k8[−32cos2x+16cos6x]+C then k=S16

A) x=t4⇒dx=4t3dt∫x14dx1+x=∫t4t3dt1+t2=4∫t4-11+t2+11+t2dt =4∫t2-1+11+t2dt=4t33-4t+4tan-1t+c B) ∫ sin((x−α)+α)sin(x−α)dx=cosα∫dx+sinα∫cot(x−α)dxC) ∫dx(x2+1)(x2+4)= 13∫1x2+1−1x2+4dxD) sin3xcos3x=18sin32x=18(3sin2x−sin6x4)

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Match the following
Column – I Column – II
A If $\int \frac{x^{\frac{1}{4}} d x}{1 + \sqrt{x}} = A x^{\frac{1}{4}} + B x^{\frac{3}{4}} + 4 \tan^{- 1} x^{\frac{1}{4}} + c,$ then $A + B$ P 1
B If $\int \frac{\sin x d x}{\sin (x - α)} = A x + B lnsin (x - α) + c,$ then $A^{2} + B^{2} =$ Q $- \frac{8}{3}$
C If $\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = A \tan^{- 1} \frac{x}{2} + B \tan^{- 1} x + c,$ then $A + B =$ R $\frac{1}{4}$
D If $\int \sin^{3} x \cos^{3} x d x = \frac{k}{8} [- \frac{3}{2} \cos 2 x + \frac{1}{6} \cos 6 x] + C$ then $k =$ S $\frac{1}{6}$

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$A \to Q; B \to P; C \to S; D \to R$

$A \to P; B \to Q; C \to R; D \to S$

$A \to S; B \to P; C \to R; D \to Q$

$A \to Q; B \to P; C \to R; D \to S$

answer is B.

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Detailed Solution

$A) x = t^{4} \Rightarrow d x = 4 t^{3} d t$

$\int \frac{x^{\frac{1}{4}} d x}{1 + \sqrt{x}} = \int \frac{t (4 t^{3}) d t}{1 + t^{2}} = 4 \int (\frac{t^{4} - 1}{1 + t^{2}} + \frac{1}{1 + t^{2}}) d t = 4 \int (t^{2} - 1 + \frac{1}{1 + t^{2}}) d t = 4 \frac{t^{3}}{3} - 4 t + 4 \tan^{- 1} t + c$

$B) \int \frac{\sin ((x - α) + α)}{\sin (x - α)} dx = \cos α \int d x + \sin α \int \cot (x - α) d x$

$C) \int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{3} \int (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4}) dx$

$D) \sin^{3} x \cos^{3} x = \frac{1}{8} \sin^{3} 2 x = \frac{1}{8} (\frac{3 \sin 2 x - \sin 6 x}{4})$

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	Column – I		Column – II
A	If $\int \frac{x^{\frac{1}{4}} d x}{1 + \sqrt{x}} = A x^{\frac{1}{4}} + B x^{\frac{3}{4}} + 4 \tan^{- 1} x^{\frac{1}{4}} + c,$ then $A + B$	P	1
B	If $\int \frac{\sin x d x}{\sin (x - α)} = A x + B lnsin (x - α) + c,$ then $A^{2} + B^{2} =$	Q	$- \frac{8}{3}$
C	If $\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = A \tan^{- 1} \frac{x}{2} + B \tan^{- 1} x + c,$ then $A + B =$	R	$\frac{1}{4}$
D	If $\int \sin^{3} x \cos^{3} x d x = \frac{k}{8} [- \frac{3}{2} \cos 2 x + \frac{1}{6} \cos 6 x] + C$ then $k =$	S	$\frac{1}{6}$

Match the following Column – I Column – II AIf ∫x14dx1+x=Ax14+Bx34+4tan−1x14+c, then A+BP1BIf ∫sinxdxsin(x−α)=Ax+Blnsin(x−α)+c, then A2+B2=Q−83CIf ∫dx(x2+1)(x2+4)=Atan−1x2+Btan−1x+c, then A+B=R14DIf ∫sin3xcos3x dx=k8[−32cos2x+16cos6x]+C then k=S16

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