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

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

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

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

- 2 ln ∣ cos (x) ∣ + cos^{2} (x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

= 2 \cdot \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

= 2 (- \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}

= 2 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = cos (x)

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

Simplify

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

= - 2 ln ∣ cos (x) ∣ + cos^{2} (x)

Add a constant to the solution

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

e^{k t} (\frac{d y}{d t} + k y) = 0

2 x^{2} y y^{^{'}} + y^{2} = 2

t an g e n t (4 x - 3)^{\frac{1}{2}}

n = 2 \sum \infty \frac{3}{n \cdot ( n + 3 )}

n \to 0 lim (\frac{1}{n} sin (\frac{1}{n}))

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