SATHEE: Indefinite Integration 1 Question 15

Indefinite Integration 1 Question 15

16. The value of $\int_{- π / 2}^{π / 2} \frac{\sin^{2} x}{1 + 2^{x}} d x$ is

(2018 Main)

(a) $\frac{π}{8}$

(b) $\frac{π}{2}$

(d) $\frac{π}{4}$

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Solution:

Key idea Use property $= \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

$\begin{array}{ll} Let & I = \int_{- π / 2}^{π / 2} \frac{\sin^{2} x}{1 + 2^{x}} d x \\ \Rightarrow & I = \int_{- π / 2}^{π / 2} \frac{\sin^{2} \frac{- π}{2} + \frac{π}{2} - x}{1 + 2^{- \frac{π}{2} + \frac{π}{2} - x}} d x \\ ∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \\ \Rightarrow & I = \int_{- π / 2}^{π / 2} \frac{\sin^{2} x}{1 + 2^{- x}} d x \\ \Rightarrow & I = \int_{- π / 2}^{π / 2} \frac{2^{x} \sin^{2} x}{2^{x} + 1} d x \\ \Rightarrow & 2 I = \int_{- π / 2}^{π / 2} \sin^{2} x \frac{2^{x} + 1}{2^{x} + 1} d x \\ \Rightarrow & 2 I = \int_{- π / 2}^{π / 2} \sin^{2} x d x \end{array}$

$\begin{array}{ll} \Rightarrow & 2 I = 2 \int_{0}^{π / 2} \sin^{2} x d x [∵ \sin^{2} x is an even function] \\ \Rightarrow & I = \int_{0}^{π / 2} \sin^{2} x d x \\ \Rightarrow & I = \int_{0}^{π / 2} \cos^{2} x d x ∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \\ \Rightarrow 2 I = \int_{0}^{π / 2} d x \\ \Rightarrow 2 I = [x]_{0}^{π / 2} \Rightarrow I = \frac{π}{4} \end{array}$

Indefinite Integration 1 Question 15

16. The value of $\int_{- π / 2}^{π / 2} \frac{\sin^{2} x}{1 + 2^{x}} d x$ is

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Indefinite Integration 1 Question 15

16. The value of ∫−π/2π/2sin2⁡x1+2xdx is

Solution:

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16. The value of $\int_{- π / 2}^{π / 2} \frac{\sin^{2} x}{1 + 2^{x}} d x$ is