What is area f(x)=sin^4(x),g(x)=sin^2(x) ?

The answer to area f(x)=sin^4(x),g(x)=sin^2(x) is (3pi)/(16)

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area f(x)=sin^4(x),g(x)=sin^2(x)

area

f (x) = sin^{4} (x), g (x) = sin^{2} (x)

\frac{3 π}{16}

Decimal Notation

0.58904 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{3 π}{2}} sin^{4} (x) - sin^{2} (x) d x

Solve

\int_{0}^{\frac{3 π}{2}} sin^{4} (x) - sin^{2} (x) d x : \frac{3 π}{16}

The area is:

= \frac{3 π}{16}

t \to 0 lim (\frac{e ^{8 t} - 1}{t})

\int e^{x} 2 x d x

\frac{d}{d x} ((\frac{2}{x} + x^{5}) (x^{3} + 1))

\frac{\partial}{\partial p} (arctan (pq r))

- cot (x) = \frac{v ^{'}}{v ^{2}}

What is area f(x)=sin^4(x),g(x)=sin^2(x) ?
The answer to area f(x)=sin^4(x),g(x)=sin^2(x) is (3pi)/(16)