What is the integral from 0 to pi/2 of 3cos^2(x) ?

The integral from 0 to pi/2 of 3cos^2(x) is (3pi)/4

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integral from 0 to pi/2 of 3cos^2(x)

\int_{0}^{\frac{π}{2}} 3 cos^{2} (x) d x

\frac{3 π}{4}

Decimal

2.35619 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 3 cos^{2} (x) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (x) d x

Rewrite using trig identities

= 3 \cdot \int_{0}^{\frac{π}{2}} \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

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 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

= 3 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d x + \int_{0}^{\frac{π}{2}} cos (2 x) d x)

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

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

= 3 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

3 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{3 π}{4}

= \frac{3 π}{4}

\frac{d}{d x} (x ln (x) + (3 x^{2} - 2)^{3})

\int 2^{3 x + 4} d x

x \to 0 lim (cos (7 x))

x \to - 1 lim (3 - 2 x)

\frac{d ^{4}}{d x ^{4}} (x)

What is the integral from 0 to pi/2 of 3cos^2(x) ?
The integral from 0 to pi/2 of 3cos^2(x) is (3pi)/4