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

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

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

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

ln ∣ tan (x) + sec (x) ∣ - sin (x) + C

Solution steps

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

Simplify

tan^{2} (x) cos (x) : sin (x) tan (x)

= \int sin (x) tan (x) d x

Rewrite using trig identities

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

Expand

\frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{cos ( x )} - cos (x)

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

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}{cos ( x )} d x - \int cos (x) d x

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= ln ∣ tan (x) + sec (x) ∣ - sin (x)

Add a constant to the solution

= ln ∣ tan (x) + sec (x) ∣ - sin (x) + C

\frac{\partial}{\partial x} (b cos (w x) + cs in (w x))

n = 0 \sum \infty (\frac{- 6 + 7}{2})^{n}

g (x) = - x^{2} + 4 x^{2} - 2 x + 2

x^{2} - 4 x, 2 x + 7

\int cos (x) + \frac{1}{9} x d x

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