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

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

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

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

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

Solution steps

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

\int \frac{x - 2}{x - 3} d x

\int \frac{x ^{4}}{( 1 + x ^{5} ) ^{\frac{2}{3}}} d x

\int \frac{2}{x 25 x ^{2} - 64} d x

\int \frac{1}{( x ^{2} + 8 x + 25 ) ^{\frac{3}{2}}} d x

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