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

The integral of (sin^3(x))/(cos^2(x)) is cos(x)+sec(x)+C

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integral of (sin^3(x))/(cos^2(x))

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

cos (x) + sec (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int 1 d u = u

= - (- \frac{1}{u}) + u

Substitute back

u = cos (x)

= - (- \frac{1}{cos ( x )}) + cos (x)

Simplify

- (- \frac{1}{cos ( x )}) + cos (x) : cos (x) + sec (x)

= cos (x) + sec (x)

Add a constant to the solution

= cos (x) + sec (x) + C

f (x) = 2 - x e^{x}, at x = 0

y^{^{'}} = e^{- y} (2 x - 4), y (0) = 1

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 t}

y = ln (x^{3} e^{2 x})

\frac{x ^{3} + 2}{3}

What is the integral of (sin^3(x))/(cos^2(x)) ?
The integral of (sin^3(x))/(cos^2(x)) is cos(x)+sec(x)+C