_{-\;π/3}^{\;π/3} ² x d x=?

Correct Answer is : π3− 34\;π/3-\;{{3}}{4}3π−43

Correct Answer is : π3− 34\;π/3-\;{{3}}{4}3π−43

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Army Group X Solved Paper

Question : 47 of 50

Marks: +1, -0

\int\limits_{-\;\frac{\pi}{3}}^{\;\frac{\pi}{3}} \sin ^{2} x d x=?

Solution:

\int\limits_{-}^{\;\frac{\pi}{3}} \sin ^{2} x d x=?

\Rightarrow \;\; ?=\int\limits_{-\;\frac{\pi}{3}}^{\;\frac{\pi}{3}} \sin ^{2} x d x

f(x)=\sin ^{2} x \Rightarrow f(-x)=\sin ^{2}(-x)

= \;\; [\sin (-x)]^{2}=(-\sin x)^{2}

\;\; =(\sin x)^{2}=\sin ^{2} x=f(x)

\therefore \;\; ? \;\; =2 \int\limits_{0}^{\;\frac{\pi}{3}} \sin ^{2} x d x

\left[ \because \int\limits_{-a}^{a} f(x) d x=2 \int\limits_{0}^{a} f(x) d x, \;\text{ if }\; f(x)=f(-x)\right]

=2 \int\limits_{0}^{\frac{\pi}{3}} \;\frac{1-\cos 2 x}{2} d x

\because \cos 2 x=1-2 \sin ^{2}

=\int\limits_{0}^{\frac{\pi}{3}} d x-\int\limits_{0}^{\frac{\pi}{3}} \cos 2 x d x

=[x]_{0}^{\frac{\pi}{3}}-\left[\;{\frac{\sin 2 x}{2}}\right]_{0}^{\frac{\pi}{3}}

=\left[\;\frac{\pi}{3}-0\right]-\;\frac{1}{2}\left[\sin \;\frac{2\pi}{3}-\sin 0\right]

=\;\frac{\pi}{3}-\;\frac{1}{2}\left(\;\frac{\sqrt{3}}{2}-0\right)

\left[ \because \sin \;\frac{2\pi}{3}=\sin 120^{\circ}=\;\frac{\sqrt{3}}{2}\right]

=\;\frac{\pi}{3}-\;\frac{\sqrt{3}}{4}

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