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

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

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

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

x - \frac{1}{2} sin (2 x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int 1 - cos (2 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

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

\int 1 d x = x

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

= 2 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

2 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : x - \frac{1}{2} sin (2 x)

= x - \frac{1}{2} sin (2 x)

Add a constant to the solution

= x - \frac{1}{2} sin (2 x) + C

(ln (x))^{4}

\frac{d ^{2}}{d x ^{2}} (cos (2 x))

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

5 y^{2} (4 y^{2} + 7)^{\frac{6}{5}}

\int - x^{2} + 1 d x

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