What is the integral of (sin^5(x))/(cos^2(x)) ?

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

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integral of (sin^5(x))/(cos^2(x))

\int \frac{sin ^{5} ( x )}{cos ^{2} ( x )} d x

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

Solution steps

\int \frac{sin ^{5} ( x )}{cos ^{2} ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int 2 d u = 2 u

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

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

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

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

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