Mathematics

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

##### SOLUTION

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

Let $\sin 2t=u$.

Then,

$d(\sin 2t)=du$

$\Rightarrow 2\cos 2t\ dt=du$

$\Rightarrow \cos 2t dt=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$

$t=\dfrac{\pi}{4}$

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

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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