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

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

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integral from pi/8 to pi/4 of 3csc^2(2x)

\int_{\frac{π}{8}}^{\frac{π}{4}} 3 csc^{2} (2 x) d x

\frac{3}{2}

Decimal

1.5

Solution steps

\int_{\frac{π}{8}}^{\frac{π}{4}} 3 csc^{2} (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_{\frac{π}{8}}^{\frac{π}{4}} csc^{2} (2 x) d x

Apply u-substitution

: \int_{\frac{π}{4}}^{\frac{π}{2}} csc^{2} (u) \frac{1}{2} d u

= 3 \cdot \int_{\frac{π}{4}}^{\frac{π}{2}} csc^{2} (u) \frac{1}{2} d u

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_{\frac{π}{4}}^{\frac{π}{2}} csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= 3 \cdot \frac{1}{2} [- cot (u)]_{\frac{π}{4}}^{\frac{π}{2}}

Simplify

3 \cdot \frac{1}{2} [- cot (u)]_{\frac{π}{4}}^{\frac{π}{2}} : \frac{3}{2} [- cot (u)]_{\frac{π}{4}}^{\frac{π}{2}}

= \frac{3}{2} [- cot (u)]_{\frac{π}{4}}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{3}{2} \cdot 1

Simplify

= \frac{3}{2}

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