Solve the integral I -0-sin 2 xdxA- -B- -2- 3 -2D- 0

The correct option is B I = ∫ sin^2 x dx =∫_0^π1 cos 2x/2 dx ∵ sin^2x=1 cos 2x/2 =1/2 [ ∫_0^π dx ∫_0^π (cos 2x)dx ]=1/2 [ x sin 2x/2 ]_0^π =1/2 [ (π 0) ...

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Byju's Answer

Standard XI

Mathematics

Solving Simultaneous Trigonometric Equations

Solve the int...

Question

Solve the integral I = $\int_{0}^{π} s i n^{2} x d x$

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Solution

The correct option is B

$I = \int s i n^{2} x d x = \int_{0}^{π} ⟮ \frac{1 - c o s 2 x}{2} ⟯ d x ⟮ ∵ s i n^{2} x = \frac{1 - c o s 2 x}{2} ⟯$

= $\frac{1}{2} [\int_{0}^{π} d x - \int_{0}^{π} (c o s 2 x) d x] = \frac{1}{2} [x - \frac{s i n 2 x}{2}]_{0}^{π}$

= $\frac{1}{2} [(π - 0) - \frac{(s i n 2 π - s i n 0)}{2}] ∴ \int_{0}^{π} s i n^{2} x d x = \frac{π}{2}$

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Solve the integral I = ∫π0sin2xdx

The correct option is B I=∫sin2xdx=∫π0⟮1−cos 2x2⟯dx⟮∵sin2x=1−cos 2x2⟯ =12[∫π0dx−∫π0(cos 2x)dx]=12[x−sin 2x2]π0 =12[(π−0)−(sin 2π−sin 0)2]∴∫π0sin2xdx=π2

Solve the integral I = $\int_{0}^{π} s i n^{2} x d x$