Mathematics

$$\int _{ 0 }^{{ \pi /2 }  }{ { x }^{ 2 } } { cos }^{ 2 }$$ x dx


SOLUTION

We have,

$$ \int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}{{\cos }^{2}}}xdx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \cos 2\theta =2{{\cos }^{2}}\theta -1 $$

$$ \Rightarrow \int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}}\dfrac{\left( 1+\cos 2x \right)}{2}dx $$

$$ \Rightarrow \int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{x}^{2}}+{{x}^{2}}\cos 2x}{2}dx} $$

$$ \Rightarrow \dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}+\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}\cos 2x}dx} $$

On integration and we get,

$$ \dfrac{1}{2}{{\left[ \dfrac{{{x}^{3}}}{3} \right]}_{0}}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ {{x}^{2}}\int_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}-\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{d}{dx}{{x}^{2}}\int_{0}^{\dfrac{\pi }{2}}{\cos 2x}dx \right)dx} \right] $$

$$ \Rightarrow \dfrac{1}{2}\left[ \dfrac{\dfrac{{{\pi }^{3}}}{8}-{{0}^{3}}}{3} \right]+\dfrac{1}{2}\left[ {{\left\{ {{x}^{2}}\left( \dfrac{\sin 2x}{2} \right) \right\}}_{0}}^{\dfrac{\pi }{2}}-\int_{0}^{\dfrac{\pi }{2}}{\left( 2x\left( \dfrac{\sin 2x}{2} \right) \right)dx} \right] $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}+\dfrac{1}{2}\left[ \left( \dfrac{\pi }{2}-0 \right)\dfrac{\left( \sin 2\times \dfrac{\pi }{2}-\sin 0 \right)}{2} \right]-\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{\left( x\sin 2x \right)dx} $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}+\dfrac{1}{2}\left[ \left( \dfrac{\pi }{2} \right)\dfrac{\left( 0-0 \right)}{2} \right]-\dfrac{1}{2}\left[ x\int_{0}^{\dfrac{\pi }{2}}{\sin 2xdx}-\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{d}{dx}x\int_{0}^{\dfrac{\pi }{2}}{\sin 2xdx} \right)dx} \right] $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}+0-\dfrac{1}{2}{{\left[ x\left( \dfrac{-\cos 2x}{2} \right) \right]}_{0}}^{\dfrac{\pi }{2}}+\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( -\cos 2x \right)}{2}dx} $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{1}{2}\left[ \left\{ \left( \dfrac{\pi }{2}-0 \right)\left( \dfrac{-\cos 2\times \dfrac{\pi }{2}+\cos 0}{2} \right) \right\} \right]-\dfrac{1}{2}{{\left[ \dfrac{\sin 2x}{2} \right]}_{0}}^{\dfrac{\pi }{2}} $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{1}{2}\left[ \left\{ \left( \dfrac{\pi }{2} \right)\left( \dfrac{1+1}{2} \right) \right\} \right]-\dfrac{1}{2}\left[ \dfrac{\sin 2\times \dfrac{\pi }{2}-\sin 0}{2} \right] $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{\pi }{4}-\dfrac{1}{2}\left[ \dfrac{0-0}{2} \right] $$

$$ \Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{\pi }{4} $$

$$ \Rightarrow \dfrac{\pi }{4}\left( \dfrac{{{\pi }^{2}}}{12}-1 \right) $$

Hence, this is the answer.
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