What is the integral of (sin^3(θ))/(cos^6(θ)) ?

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

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integral of (sin^3(θ))/(cos^6(θ))

\int \frac{sin ^{3} ( θ )}{cos ^{6} ( θ )} d θ

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

Solution steps

\int \frac{sin ^{3} ( θ )}{cos ^{6} ( θ )} d θ

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u ^{6}} d u

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

Expand

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

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

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

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

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

Substitute back

u = cos (θ)

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

Simplify

- (- \frac{1}{5 cos ^{5} ( θ )}) - \frac{1}{3 cos ^{3} ( θ )} : \frac{1}{5} sec^{5} (θ) - \frac{1}{3} sec^{3} (θ)

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

Add a constant to the solution

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

\int \frac{sin ( 2 x )}{2} d x

\frac{\partial}{\partial x} (x^{z} y^{1 - z})

- 3 e^{x + 4} + e^{3}

f (x) = (x^{2} - 18) (2 x - 3)

\frac{d}{d x} (\frac{- x ^{3}}{y ^{3}})

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