What is the integral of (sin^5(x))/(cos^6(x)) ?

The integral of (sin^5(x))/(cos^6(x)) is 1/5 sec^5(x)-2/3 sec^3(x)+sec(x)+C

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integral of (sin^5(x))/(cos^6(x))

\int \frac{sin ^{5} ( x )}{cos ^{6} ( x )} d x

\frac{1}{5} sec^{5} (x) - \frac{2}{3} sec^{3} (x) + sec (x) + C

Solution steps

\int \frac{sin ^{5} ( x )}{cos ^{6} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) ^{2} sin ( x )}{cos ^{6} ( x )} d x

Apply u-substitution

= \int - \frac{( 1 - u ^{2} ) ^{2}}{u ^{6}} d u

Expand

- \frac{( 1 - u ^{2} ) ^{2}}{u ^{6}} : - \frac{1}{u ^{6}} + \frac{2}{u ^{4}} - \frac{1}{u ^{2}}

= \int - \frac{1}{u ^{6}} + \frac{2}{u ^{4}} - \frac{1}{u ^{2}} 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 ^{6}} d u + \int \frac{2}{u ^{4}} d u - \int \frac{1}{u ^{2}} d u

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

\int \frac{2}{u ^{4}} d u = - \frac{2}{3 u ^{3}}

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

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

Substitute back

u = cos (x)

= - (- \frac{1}{5 cos ^{5} ( x )}) - \frac{2}{3 cos ^{3} ( x )} - (- \frac{1}{cos ( x )})

Simplify

- (- \frac{1}{5 cos ^{5} ( x )}) - \frac{2}{3 cos ^{3} ( x )} - (- \frac{1}{cos ( x )}) : \frac{1}{5} sec^{5} (x) - \frac{2}{3} sec^{3} (x) + sec (x)

= \frac{1}{5} sec^{5} (x) - \frac{2}{3} sec^{3} (x) + sec (x)

Add a constant to the solution

= \frac{1}{5} sec^{5} (x) - \frac{2}{3} sec^{3} (x) + sec (x) + C

What is the integral of (sin^5(x))/(cos^6(x)) ?
The integral of (sin^5(x))/(cos^6(x)) is 1/5 sec^5(x)-2/3 sec^3(x)+sec(x)+C