What is the integral of cos^2(x)tan^3(x) ?

The integral of cos^2(x)tan^3(x) is -ln|cos(x)|+(cos^2(x))/2+C

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integral of cos^2(x)tan^3(x)

\int cos^{2} (x) tan^{3} (x) d x

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

Solution steps

\int cos^{2} (x) tan^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = cos (x)

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

Add a constant to the solution

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

f (x) = \frac{e ^{x}}{x ^{2}}

x \to \frac{1}{4} lim (8 x (x - \frac{1}{5}))

x \to 0 lim (\frac{1}{1 - x})

y^{^{'}} = 0.5 (3 - y)

y^{^{''}} - 21 y^{^{'}} + 108 y = 0

What is the integral of cos^2(x)tan^3(x) ?
The integral of cos^2(x)tan^3(x) is -ln|cos(x)|+(cos^2(x))/2+C