Mathematics

Evaluate: $$\displaystyle \int\limits_0^{\frac{\pi }{4}} {{{\cos }^{\frac{3}{2}}}\left( {2x} \right)\cos \left( x \right)dx} $$


SOLUTION
$$\int_0^{\frac{\pi}{4}}(\cos 2x)^{\frac{3}{2}}\cos x dx$$
$$\Rightarrow$$  $$\int_0^{\frac{\pi}{4}}(1-2\sin^2x)^{\frac{3}{2}}\cos x dx$$
Let $$\sqrt{2}\sin x=\sin t$$
$$\cos x dx=\dfrac{1}{\sqrt{2}}\cos tdt$$

$$\Rightarrow$$  $$\dfrac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^4tdt$$      ---- ( 1 )
Now, $$\int \cos^4tdt$$
  $$=\dfrac{1}{4}\int (2\cos^2t)dt$$
  $$=\dfrac{1}{4}\int(1+\cos 2t)^2dt$$
  $$=\dfrac{1}{4}\int[1+2\cos 2t+\cos^2 2t]dt$$
  $$=\dfrac{1}{4}\left[t+\sin 2t+\dfrac{1}{2}\int(1+\cos 4t)dt\right]$$
  $$=\dfrac{1}{4}\left[t+\sin 2t+\dfrac{1}{2}t+\dfrac{1}{8}\sin 4t\right]_0^{\frac{\pi}{2}}+c$$
  $$=\dfrac{1}{4}\left[\dfrac{3}{2}t+\sin 2t+\dfrac{1}{8}\sin 4t\right]_0^{\frac{\pi}{2}}+c$$         ---- ( 2 )
Substituting ( 2 ) in ( 1 )
$$\Rightarrow$$  $$\dfrac{1}{4\sqrt{2}}\left[\dfrac{3}{2}t+\sin 2t+\dfrac{1}{8}\sin 4t\right]_0^{\frac{\pi}{2}}+c$$  

$$\Rightarrow$$  $$\dfrac{1}{4\sqrt{2}}\left[\left(\dfrac{3}{2}.\dfrac{\pi}{2}-0\right)+(0-0)+\dfrac{1}{8}(0-0)\right]$$

$$\Rightarrow$$  $$\dfrac{3\pi}{16\sqrt{2}}$$


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Subjective Medium Published on 17th 09, 2020
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