Mathematics

Evaluate the integral $$\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$$.


SOLUTION
Let $$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$$ 

and let $$\sin 2t=u$$. Then, 

$$d(\sin 2t)=du\Rightarrow 2\cos 2t\ dt=du\Rightarrow \cos 2t dt=\dfrac{1}{2}du$$

Also, $$t=0\Rightarrow u=\sin 0=0$$ 

and $$t=\dfrac{\pi}{4}\Rightarrow u=\sin\dfrac{\pi}{2}=1$$

$$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du$$ 

$$=\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}$$ 

$$=\dfrac{1}{8}$$

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Subjective Medium Published on 17th 09, 2020
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