What is the integral of sin^3(4x)cos^{-2}(4x) ?

The integral of sin^3(4x)cos^{-2}(4x) is 1/4 (sec(4x)+cos(4x))+C

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integral of sin^3(4x)cos^{-2}(4x)

\int sin^{3} (4 x) cos^{- 2} (4 x) d x

\frac{1}{4} (sec (4 x) + cos (4 x)) + C

Solution steps

\int sin^{3} (4 x) cos^{- 2} (4 x) d x

Apply u-substitution

: \int \frac{sin ^{3} ( u )}{4 cos ^{2} ( u )} d u

= \int \frac{sin ^{3} ( u )}{4 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sin ^{3} ( u )}{cos ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{( 1 - cos ^{2} ( u ) ) sin ( u )}{cos ^{2} ( u )} d u

Apply u-substitution

: \int - \frac{1 - v ^{2}}{v ^{2}} d v

= \frac{1}{4} \cdot \int - \frac{1 - v ^{2}}{v ^{2}} d v

Expand

- \frac{1 - v ^{2}}{v ^{2}} : - \frac{1}{v ^{2}} + 1

= \frac{1}{4} \cdot \int - \frac{1}{v ^{2}} + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (- \int \frac{1}{v ^{2}} d v + \int 1 d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int 1 d v = v

= \frac{1}{4} (- (- \frac{1}{v}) + v)

Substitute back

= \frac{1}{4} (- (- \frac{1}{cos ( 4 x )}) + cos (4 x))

Simplify

\frac{1}{4} (- (- \frac{1}{cos ( 4 x )}) + cos (4 x)) : \frac{1}{4} (sec (4 x) + cos (4 x))

= \frac{1}{4} (sec (4 x) + cos (4 x))

Add a constant to the solution

= \frac{1}{4} (sec (4 x) + cos (4 x)) + C

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What is the integral of sin^3(4x)cos^{-2}(4x) ?
The integral of sin^3(4x)cos^{-2}(4x) is 1/4 (sec(4x)+cos(4x))+C