What is the integral of sin(3x)cos^{-4}(3x) ?

The integral of sin(3x)cos^{-4}(3x) is 1/9 sec^3(3x)+C

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integral of sin(3x)cos^{-4}(3x)

\int sin (3 x) cos^{- 4} (3 x) d x

\frac{1}{9} sec^{3} (3 x) + C

Solution steps

\int sin (3 x) cos^{- 4} (3 x) d x

Apply u-substitution

: \int \frac{sin ( u )}{3 cos ^{4} ( u )} d u

= \int \frac{sin ( u )}{3 cos ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sin ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{tan ( u )}{cos ^{3} ( u )} d u

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= \frac{1}{3} \cdot \int sec^{3} (u) tan (u) d u

Apply u-substitution

: \int \frac{1}{3} d v

= \frac{1}{3} \cdot \int \frac{1}{3} d v

Integral of a constant:

\int a d x = a x

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

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} sec^{3} (3 x)

Simplify

\frac{1}{3} \cdot \frac{1}{3} sec^{3} (3 x) : \frac{1}{9} sec^{3} (3 x)

= \frac{1}{9} sec^{3} (3 x)

Add a constant to the solution

= \frac{1}{9} sec^{3} (3 x) + C

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What is the integral of sin(3x)cos^{-4}(3x) ?
The integral of sin(3x)cos^{-4}(3x) is 1/9 sec^3(3x)+C