What is the integral from 0 to pi/6 of cos^{-3}(2x)sin(2x) ?

The integral from 0 to pi/6 of cos^{-3}(2x)sin(2x) is 3/4

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integral from 0 to pi/6 of cos^{-3}(2x)sin(2x)

\int_{0}^{\frac{π}{6}} cos^{- 3} (2 x) sin (2 x) d x

\frac{3}{4}

Decimal

0.75

Solution steps

\int_{0}^{\frac{π}{6}} cos^{- 3} (2 x) sin (2 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} \frac{2 cos ( x ) sin ( x )}{cos ^{3} ( 2 x )} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{π}{6}} \frac{cos ( x ) sin ( x )}{cos ^{3} ( 2 x )} d x

Rewrite using trig identities

= 2 \cdot \int_{0}^{\frac{π}{6}} \frac{sin ( 2 x )}{2 cos ^{3} ( 2 x )} d x

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} \frac{sin ( 2 x )}{cos ^{3} ( 2 x )} d x

Apply u-substitution

: \int_{0}^{\frac{π}{3}} \frac{sin ( u )}{2 cos ^{3} ( u )} d u

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

Take the constant out:

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

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

Rewrite using trig identities

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} tan (u) sec^{2} (u) d u

Apply u-substitution

: \int_{0}^{3} v d v

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

Apply the Power Rule

: [\frac{v ^{2}}{2}]_{0}^{3}

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3} : \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

= \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

Compute the boundaries:

\frac{3}{2}

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

Simplify

= \frac{3}{4}

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What is the integral from 0 to pi/6 of cos^{-3}(2x)sin(2x) ?
The integral from 0 to pi/6 of cos^{-3}(2x)sin(2x) is 3/4