What is the integral of (sin^5(x))/(cos(x)) ?

The integral of (sin^5(x))/(cos(x)) is -ln|cos(x)|+cos^2(x)-(cos^4(x))/4+C

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integral of (sin^5(x))/(cos(x))

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

- ln ∣ cos (x) ∣ + cos^{2} (x) - \frac{cos ^{4} ( x )}{4} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{( 1 - u ^{2} ) ^{2}}{u} d u

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

Expand

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

= \int - \frac{1}{u} + 2 u - u^{3} 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} d u + \int 2 u d u - \int u^{3} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int u^{3} d u = \frac{u ^{4}}{4}

= - ln ∣ u ∣ + u^{2} - \frac{u ^{4}}{4}

Substitute back

u = cos (x)

= - ln ∣ cos (x) ∣ + cos^{2} (x) - \frac{cos ^{4} ( x )}{4}

Add a constant to the solution

= - ln ∣ cos (x) ∣ + cos^{2} (x) - \frac{cos ^{4} ( x )}{4} + C

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What is the integral of (sin^5(x))/(cos(x)) ?
The integral of (sin^5(x))/(cos(x)) is -ln|cos(x)|+cos^2(x)-(cos^4(x))/4+C