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Sophie [7]
3 years ago
11

Integration of sin^2 2x cos^3 2x dx

Mathematics
1 answer:
lilavasa [31]3 years ago
4 0

\int\left[\sin^2(2x)\cos^3(2x)\right]dx=\int\left[\sin^2(2x)\cos^2(2x)\cos(2x)\right]dx\\\\\text{Use the trigonometric identity}\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\=\int\left\{\sin^2(2x)[1-\sin^2(2x)]\cos(2x)\right\}dx\\\\\text{substitute}\ \sin2x=t\ \to\ 2\cos2x\ dx=dt\ \to\ \cos2x\ dx=\dfrac{1}{2}\ dt\\\\=\dfrac{1}{2}\int\left[t^2(1-t^2)\right]dt=\dfrac{1}{2}\int(t^2-t^4)dt=\dfrac{1}{2}\left(\dfrac{1}{3}t^3-\dfrac{1}{5}t^5\right)+C


=\dfrac{1}{2}\left(\dfrac{1}{3}\sin^3(2x)-\dfrac{1}{5}\sin^5(2x)\right)+C

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Answer:

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Step-by-step explanation:

<u>Given:</u>

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<u>Find the future amount:</u>

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Step-by-step explanation:

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Hence, base = 2 units and hypotenuse = 3 units.

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\quad \twoheadrightarrow\sf { P^2 = H^2 - B^2} \\

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Now, we know that :

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