What is the integral of sin^{-4}(x/5)cos^3(x/5) ?

The integral of sin^{-4}(x/5)cos^3(x/5) is 5(-1/3 csc^3(x/5)+csc(x/5))+C

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integral of sin^{-4}(x/5)cos^3(x/5)

\int sin^{- 4} (\frac{x}{5}) cos^{3} (\frac{x}{5}) d x

5 (- \frac{1}{3} csc^{3} (\frac{x}{5}) + csc (\frac{x}{5})) + C

Solution steps

\int sin^{- 4} (\frac{x}{5}) cos^{3} (\frac{x}{5}) d x

Apply u-substitution

: \int \frac{5 cos ^{3} ( u )}{sin ^{4} ( u )} d u

= \int \frac{5 cos ^{3} ( u )}{sin ^{4} ( u )} d u

Take the constant out:

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

= 5 \cdot \int \frac{cos ^{3} ( u )}{sin ^{4} ( u )} d u

Rewrite using trig identities

= 5 \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

= 5 \cdot \int \frac{1 - v ^{2}}{v ^{4}} d v

Expand

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

= 5 \cdot \int \frac{1}{v ^{4}} - \frac{1}{v ^{2}} 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

= 5 (\int \frac{1}{v ^{4}} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

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

= 5 (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

= 5 (- \frac{1}{3 sin ^{3} ( \frac{x}{5} )} - (- \frac{1}{sin ( \frac{x}{5} )}))

Simplify

5 (- \frac{1}{3 sin ^{3} ( \frac{x}{5} )} - (- \frac{1}{sin ( \frac{x}{5} )})) : 5 (- \frac{1}{3} csc^{3} (\frac{x}{5}) + csc (\frac{x}{5}))

= 5 (- \frac{1}{3} csc^{3} (\frac{x}{5}) + csc (\frac{x}{5}))

Add a constant to the solution

= 5 (- \frac{1}{3} csc^{3} (\frac{x}{5}) + csc (\frac{x}{5})) + C

f (x) = (0.46) + (- 4.39 x) + (- 206 x^{2}) + (2.8 x^{3})

x \to - 1 lim (\frac{x ^{2}}{x ^{2} - 1})

\frac{d}{d x} (cos (20 x))

7 + 5 x + x e^{x}

x \to 8 + lim (\frac{x - 9}{x - 8})

What is the integral of sin^{-4}(x/5)cos^3(x/5) ?
The integral of sin^{-4}(x/5)cos^3(x/5) is 5(-1/3 csc^3(x/5)+csc(x/5))+C