What is the integral of sin^{-5}(x)cos^7(x) ?

The integral of sin^{-5}(x)cos^7(x) is -1/4 csc^4(x)+3/2 csc^2(x)+3ln|sin(x)|-1/2 sin^2(x)+C

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integral of sin^{-5}(x)cos^7(x)

\int sin^{- 5} (x) cos^{7} (x) d x

- \frac{1}{4} csc^{4} (x) + \frac{3}{2} csc^{2} (x) + 3 ln ∣ sin (x) ∣ - \frac{1}{2} sin^{2} (x) + C

Solution steps

\int sin^{- 5} (x) cos^{7} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x))^{3} cos (x) sin^{- 5} (x) d x

Apply u-substitution

= \int \frac{( 1 - u ^{2} ) ^{3}}{u ^{5}} d u

Expand

\frac{( 1 - u ^{2} ) ^{3}}{u ^{5}} : \frac{1}{u ^{5}} - \frac{3}{u ^{3}} + \frac{3}{u} - u

= \int \frac{1}{u ^{5}} - \frac{3}{u ^{3}} + \frac{3}{u} - u d u

Apply the Sum Rule:

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

= \int \frac{1}{u ^{5}} d u - \int \frac{3}{u ^{3}} d u + \int \frac{3}{u} d u - \int u d u

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

\int \frac{3}{u ^{3}} d u = - \frac{3}{2 u ^{2}}

\int \frac{3}{u} d u = 3 ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= - \frac{1}{4 u ^{4}} - (- \frac{3}{2 u ^{2}}) + 3 ln ∣ u ∣ - \frac{u ^{2}}{2}

Substitute back

u = sin (x)

= - \frac{1}{4 sin ^{4} ( x )} - (- \frac{3}{2 sin ^{2} ( x )}) + 3 ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2}

Simplify

- \frac{1}{4 sin ^{4} ( x )} - (- \frac{3}{2 sin ^{2} ( x )}) + 3 ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2} : - \frac{1}{4} csc^{4} (x) + \frac{3}{2} csc^{2} (x) + 3 ln ∣ sin (x) ∣ - \frac{1}{2} sin^{2} (x)

= - \frac{1}{4} csc^{4} (x) + \frac{3}{2} csc^{2} (x) + 3 ln ∣ sin (x) ∣ - \frac{1}{2} sin^{2} (x)

Add a constant to the solution

= - \frac{1}{4} csc^{4} (x) + \frac{3}{2} csc^{2} (x) + 3 ln ∣ sin (x) ∣ - \frac{1}{2} sin^{2} (x) + C

What is the integral of sin^{-5}(x)cos^7(x) ?
The integral of sin^{-5}(x)cos^7(x) is -1/4 csc^4(x)+3/2 csc^2(x)+3ln|sin(x)|-1/2 sin^2(x)+C