What is the integral from-pi/2 to pi/2 of sin^4(x) ?

The integral from-pi/2 to pi/2 of sin^4(x) is (3pi)/8

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integral from-pi/2 to pi/2 of sin^4(x)

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

\frac{3 π}{8}

Decimal

1.17809 \dots

Solution steps

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

Apply Integral Reduction

= - [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} sin^{2} (x) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} sin^{2} (x) d x = \frac{π}{2}

= - [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{2}

Simplify

- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{2} : - [\frac{1}{4} sin^{3} (x) cos (x)]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{3 π}{8}

= - [\frac{1}{4} sin^{3} (x) cos (x)]_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{3 π}{8}

Compute the boundaries:

0

= - 0 + \frac{3 π}{8}

Simplify

= \frac{3 π}{8}

\int_{1}^{2} (\frac{x}{x ^{2} - 3}) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} 3 cos (x) d x

\int_{0}^{\infty} e^{- a t} d t

\int_{- 1}^{2} x - x^{2} d x

\int_{0}^{y} x e^{- x} d x

What is the integral from-pi/2 to pi/2 of sin^4(x) ?
The integral from-pi/2 to pi/2 of sin^4(x) is (3pi)/8