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

The integral from 0 to pi/4 of (3cos(2x))^2 is (9pi)/8

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

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

\frac{9 π}{8}

Decimal

3.53429 \dots

Solution steps

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

Apply Integration By Parts

: [9 x cos^{2} (2 x) - \int - 36 x cos (2 x) sin (2 x) d x]_{0}^{\frac{π}{4}}

= [9 x cos^{2} (2 x) - \int - 36 x cos (2 x) sin (2 x) d x]_{0}^{\frac{π}{4}}

\int - 36 x cos (2 x) sin (2 x) d x = - 36 (- \frac{1}{8} x cos (4 x) + \frac{1}{32} sin (4 x))

= [9 x cos^{2} (2 x) - (- 36 (- \frac{1}{8} x cos (4 x) + \frac{1}{32} sin (4 x)))]_{0}^{\frac{π}{4}}

Simplify

= [9 x cos^{2} (2 x) + 36 (- \frac{1}{8} x cos (4 x) + \frac{1}{32} sin (4 x))]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{9 π}{8}

= \frac{9 π}{8}

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\int_{0}^{4} x^{2} e^{\frac{x}{2}} d x

\int_{5}^{8} 2400 x 3 + 2 x d x

What is the integral from 0 to pi/4 of (3cos(2x))^2 ?
The integral from 0 to pi/4 of (3cos(2x))^2 is (9pi)/8