Match the definite integrals in Column – I with their values in Column – II. Column – I Column – II A)∫0π2f(sin2x).sinx dxP)π∫0π2f(cosx) dxB)∫0π2f(sin2x).cosx dxQ)∫0π42f(cos2x).cosx dxC)∫0πx.f(sinx) dxR)12∫7π49π4f(cos2x).cosx dxD)∫0π2x.f(sin2x) dxS)π4∫0π2f(cosx) dx

A) I=∫0π2f(sin2x)sinxdx ..........(1) =∫0π2f(sinπ−2x)cosxdx =∫0π2f(sin2x)cosxdx..........(2) 2I=∫0π2f(sin2x){sinx+cosx}dx I=12∫0π2f(sin2x)2cos(x−π4)dx=12∫0π2f(sin2x)cos(x−π4)dx let x−π4=t⇒dx=dt I=12∫−(π4)π4f(cos2t)costdt=2∫0π4f(cos2t)costdt =12∫2π−(π/4)2π+(π/4)f(cos2t)costdt=2∫7π49π4f(cos2t)costdt [Since 2π is Period]C) I=∫0πxf(sinx)dx⇒I=∫0π(π−x)(sinx)dx ⇒2I=∫0πxf(sinx)dx⇒I=π2∫0πf(sinx)dx ⇒I=∫0π2f(sinx)dx=∫0π2f(cosx)dxD) ⇒I=∫0π2x.f(sin2x)dx=∫0π2(π2−x)f(sin2x)dx ⇒2I=∫0π2π2f(sin2x)dx⇒I=∫0π2π4f(sin2x)dx put 2x=t I=∫0ππ4f(sint)dt2=π8∫0πf(sint)dt =π4∫0π2f(sint)dt=π4∫0π2f(cost)dt

Match the definite integrals in Column – I with their values in Column – II. Column – I Column – II A)∫0π2f(sin2x).sinx dxP)π∫0π2f(cosx) dxB)∫0π2f(sin2x).cosx dxQ)∫0π42f(cos2x).cosx dxC)∫0πx.f(sinx) dxR)12∫7π49π4f(cos2x).cosx dxD)∫0π2x.f(sin2x) dxS)π4∫0π2f(cosx) dx

A) I=∫0π2f(sin2x)sinxdx ..........(1) =∫0π2f(sinπ−2x)cosxdx =∫0π2f(sin2x)cosxdx..........(2) 2I=∫0π2f(sin2x){sinx+cosx}dx I=12∫0π2f(sin2x)2cos(x−π4)dx=12∫0π2f(sin2x)cos(x−π4)dx let x−π4=t⇒dx=dt I=12∫−(π4)π4f(cos2t)costdt=2∫0π4f(cos2t)costdt =12∫2π−(π/4)2π+(π/4)f(cos2t)costdt=2∫7π49π4f(cos2t)costdt [Since 2π is Period]C) I=∫0πxf(sinx)dx⇒I=∫0π(π−x)(sinx)dx ⇒2I=∫0πxf(sinx)dx⇒I=π2∫0πf(sinx)dx ⇒I=∫0π2f(sinx)dx=∫0π2f(cosx)dxD) ⇒I=∫0π2x.f(sin2x)dx=∫0π2(π2−x)f(sin2x)dx ⇒2I=∫0π2π2f(sin2x)dx⇒I=∫0π2π4f(sin2x)dx put 2x=t I=∫0ππ4f(sint)dt2=π8∫0πf(sint)dt =π4∫0π2f(sint)dt=π4∫0π2f(cost)dt

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Match the definite integrals in Column – I with their values in Column – II.
Column – I Column – II
A) $\int_{0}^{\frac{π}{2}} f (\sin 2 x) . \sin x d x$ P) $π \int_{0}^{\frac{π}{2}} f (\cos x) d x$
B) $\int_{0}^{\frac{π}{2}} f (\sin 2 x) . \cos x d x$ Q) $\int_{0}^{\frac{π}{4}} \sqrt{2} f (\cos 2 x) . \cos x d x$
C) $\int_{0}^{π} x . f (\sin x) d x$ R) $\frac{1}{\sqrt{2}} \int_{\frac{7 π}{4}}^{\frac{9 π}{4}} f (\cos 2 x) . \cos x d x$
D) $\int_{0}^{\frac{π}{2}} x . f (\sin 2 x) d x$ S) $\frac{π}{4} \int_{0}^{\frac{π}{2}} f (\cos x) d x$

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$A - P Q; B - Q R; C - S; D - R$

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Detailed Solution

$A) I = \int_{0}^{\frac{π}{2}} f (\sin 2 x) \sin x d x .......... (1) = \int_{0}^{\frac{π}{2}} f (\sin (π - 2 x)) \cos x d x = \int_{0}^{\frac{π}{2}} f (\sin 2 x) \cos x d x .......... (2) 2 I = \int_{0}^{\frac{π}{2}} f (\sin 2 x) {\sin x + \cos x} d x I = \frac{1}{2} \int_{0}^{\frac{π}{2}} f (\sin 2 x) \sqrt{2} \cos (x - \frac{π}{4}) d x = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} f (\sin 2 x) \cos (x - \frac{π}{4}) d x l e t x - \frac{π}{4} = t \Rightarrow d x = d t I = \frac{1}{\sqrt{2}} \int_{- (\frac{π}{4})}^{\frac{π}{4}} f (\cos 2 t) \cos t d t = \sqrt{2} \int_{0}^{\frac{π}{4}} f (\cos 2 t) \cos t d t = \frac{1}{\sqrt{2}} \int_{2 π - (π / 4)}^{2 π + (π / 4)} f (\cos 2 t) \cos t d t = \sqrt{2} \int_{\frac{7 π}{4}}^{\frac{9 π}{4}} f (\cos 2 t) \cos t d t [Since 2 π is Period]$

$C) I = \int_{0}^{π} x f (\sin x) d x \Rightarrow I = \int_{0}^{π} (π - x) (\sin x) d x \Rightarrow 2 I = \int_{0}^{π} x f (\sin x) d x \Rightarrow I = \frac{π}{2} \int_{0}^{π} f (\sin x) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} f (\sin x) d x = \int_{0}^{\frac{π}{2}} f (\cos x) d x$

$D) \Rightarrow I = \int_{0}^{\frac{π}{2}} x . f (\sin 2 x) d x = \int_{0}^{\frac{π}{2}} (\frac{π}{2} - x) f (\sin 2 x) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \frac{π}{2} f (\sin 2 x) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{π}{4} f (\sin 2 x) d x p u t 2 x = t I = \int_{0}^{π} \frac{π}{4} f (\sin t) \frac{d t}{2} = \frac{π}{8} \int_{0}^{π} f (\sin t) d t = \frac{π}{4} \int_{0}^{\frac{π}{2}} f (\sin t) d t = \frac{π}{4} \int_{0}^{\frac{π}{2}} f (\cos t) d t$

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