What is the integral from pi to 0 of sin^4(x) ?

The integral from pi to 0 of sin^4(x) is -(3pi)/8

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integral from pi to 0 of sin^4(x)

\int_{π}^{0} sin^{4} (x) d x

- \frac{3 π}{8}

Decimal

- 1.17809 \dots

Solution steps

\int_{π}^{0} sin^{4} (x) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= - (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (x) d x)

\int_{0}^{π} sin^{2} (x) d x = \frac{π}{2}

= - (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

- (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} - \frac{3 π}{8}

= [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} - \frac{3 π}{8}

Compute the boundaries:

0

= 0 - \frac{3 π}{8}

Simplify

= - \frac{3 π}{8}

\int_{1}^{2} (- 4 t^{- 4}) d t

\int_{x}^{x^{2}} e^{- t^{2}} d t

\int_{1}^{6} \frac{3 ( ln ( x ) ) ^{2}}{x} d x

\int_{2}^{x} t d t

\int_{0}^{\frac{π}{4}} - cos (x) d x

What is the integral from pi to 0 of sin^4(x) ?
The integral from pi to 0 of sin^4(x) is -(3pi)/8