-0-4sin 22 x dx is equal toa -2c -8b -4d 0

I=∫_0^π/4sin^2 (2x) dx nbsp; Using property of integration nbsp; I=∫_0^π/4sin^2 (π/2 x ) dx ...(i) ⇒ I=∫_0^π/4cos^2 (π/2 x ) dx ...(ii) Adding (i) a ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

∫0π/4sin 22 x...

Question

$\frac{π}{4} \int 0 {sin}^{2} (2 x) d x$ is equal to

$(a) \frac{π}{2} (b) \frac{π}{4}$
$(c) \frac{π}{8} (d) 0$

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Solution

$I = \frac{π}{4} \int 0 {sin}^{2} (2 x) d x$
Using property of integration
$I = \frac{π}{4} \int 0 {sin}^{2} (\frac{π}{2} - x) d x . . . (i)$
$\Rightarrow I = \frac{π}{4} \int 0 {cos}^{2} (\frac{π}{2} - x) d x . . . (i i)$

Adding $(i)$ and $(i i)$ , we get:
$2 I = \frac{π}{4} \int 0 1 d x = \frac{π}{4}$
$\Rightarrow I = \frac{π}{8}$

Option $(c)$ is correct.

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π4∫0sin2(2x) dx is equal to (a) π2 (b) π4 (c) π8 (d) 0

I=π4∫0sin2(2x) dx Using property of integration I=π4∫0sin2(π2−x) dx ...(i) ⇒I=π4∫0cos2(π2−x) dx ...(ii) Adding (i) and (ii), we get: 2I=π4∫0 1 dx=π4 ⇒I=π8 Option (c) is correct.

$\frac{π}{4} \int 0 {sin}^{2} (2 x) d x$ is equal to

$(a) \frac{π}{2} (b) \frac{π}{4}$
$(c) \frac{π}{8} (d) 0$