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

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

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

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

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

Solution steps

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

Simplify

cos (x) tan^{2} (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

y^{^{'}} = - 2 y, y (0) = 2

\frac{\partial}{\partial y} (x y e^{5 y})

x = 6.4 cos (1. 2^{2} t^{2})

\int sin (4 x) sin (3 x) d x

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