What is the integral from 0 to pi of 8cos^2(x) ?

The integral from 0 to pi of 8cos^2(x) is 4pi

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integral from 0 to pi of 8cos^2(x)

\int_{0}^{π} 8 cos^{2} (x) d x

4 π

Decimal

12.56637 \dots

Solution steps

\int_{0}^{π} 8 cos^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{0}^{π} cos^{2} (x) d x

Rewrite using trig identities

= 8 \cdot \int_{0}^{π} \frac{1 + cos ( 2 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 + cos (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 \cdot \frac{1}{2} (\int_{0}^{π} 1 d x + \int_{0}^{π} cos (2 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (2 x) d x = 0

= 8 \cdot \frac{1}{2} (π + 0)

Simplify

8 \cdot \frac{1}{2} (π + 0) : 4 π

= 4 π

\int_{0}^{2} \frac{1}{2 x - 1} d x

\int_{0}^{3} ∣ 3 t - 8 ∣ d t

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

\int_{0}^{6} \frac{7}{8 x ^{2} + 9 x + 1} d x

\int_{- 1}^{1} \frac{x}{1 + ∣ x ∣} d x

What is the integral from 0 to pi of 8cos^2(x) ?
The integral from 0 to pi of 8cos^2(x) is 4pi