If I- - 2- - 2sin 2 x-1-4x d x- then the value of 24- I is

I=∫_ π/2^π/2sin^2 x1+4^x dx ⋯ (1) ⇒ I=∫_ π/2^π/2sin^2 ( x)1+4^ x dx ⇒ I= ∫_ π/2^π/2sin^2 x1+4^ x dx ⇒ I= ∫_ π/2^π/24^xsin^2 x1+4^x dx ⋯ (2) Adding (1) ...

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Byju's Answer

Standard XII

Mathematics

Property 4

If I=∫-π / 2π...

Question

If $I = π / 2 \int - π / 2 \frac{{sin}^{2} x}{1 + 4^{x}} d x,$ then the value of $\frac{24}{π} I$ is

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Solution

$I = π / 2 \int - π / 2 \frac{{sin}^{2} x}{1 + 4^{x}} d x \dots (1)$
$\Rightarrow I = π / 2 \int - π / 2 \frac{{sin}^{2} (- x)}{1 + 4^{- x}} d x$
$\Rightarrow I = π / 2 \int - π / 2 \frac{{sin}^{2} x}{1 + 4^{- x}} d x$
$\Rightarrow I = π / 2 \int - π / 2 \frac{4^{x} {sin}^{2} x}{1 + 4^{x}} d x \dots (2)$
Adding $(1)$ and $(2),$ we get
$2 I = π / 2 \int - π / 2 {sin}^{2} x d x$
$\Rightarrow 2 I = 2 π / 2 \int 0 {sin}^{2} x d x$
$\Rightarrow I = π / 2 \int 0 {sin}^{2} x d x = \frac{π}{4}$

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If I=π/2∫−π/2sin2x1+4x dx, then the value of 24πI is

I=π/2∫−π/2sin2x1+4x dx ⋯(1) ⇒I=π/2∫−π/2sin2(−x)1+4−x dx ⇒I=π/2∫−π/2sin2x1+4−x dx ⇒I=π/2∫−π/24xsin2x1+4x dx ⋯(2) Adding (1) and (2), we get 2I=π/2∫−π/2sin2x dx ⇒2I=2π/2∫0sin2x dx ⇒I=π/2∫0sin2x dx=π4

If $I = π / 2 \int - π / 2 \frac{{sin}^{2} x}{1 + 4^{x}} d x,$ then the value of $\frac{24}{π} I$ is