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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) ^{2} sin ( x )}{cos ^{3} ( x )} d x

Apply u-substitution

: \int - \frac{( 1 - u ^{2} ) ^{2}}{u ^{3}} d u

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

Expand

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

= \int - \frac{1}{u ^{3}} + \frac{2}{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 ^{3}} d u + \int \frac{2}{u} d u - \int u d u

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

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

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

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

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