I=∫_ 1^1 3x^2 dx/1+4^tan x I=∫_ 1^1 3x^2 dx.4^tan x/4^tan x+1{∵∫_a^b f(x)dx= ∫_a^b f(a+b x)dx} ∴ 2I=∫_ 1^1 3x^2 ⇒ I=1 ...

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Integration of Piecewise Continuous Functions

∫-11 3x2 dx/1...

Question

$\int_{- 1}^{1} \frac{3 x^{2} d x}{1 + 4^{t a n x}} =$

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Solution

The correct option is C 1
$I = \int_{- 1}^{1} \frac{3 x^{2} d x}{1 + 4^{t a n x}}$
$I = \int_{- 1}^{1} \frac{3 x^{2} d x {.4}^{t a n x}}{4^{t a n x} + 1} {∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x}$
$∴ 2 I = \int_{- 1}^{1} 3 x^{2} \Rightarrow I = 1$

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