What is the integral of (tan^2(x)+cot^2(x)+4) ?

The integral of (tan^2(x)+cot^2(x)+4) is 2x+tan(x)-cot(x)+C

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integral of (tan^2(x)+cot^2(x)+4)

\int (tan^{2} (x) + cot^{2} (x) + 4) d x

2 x + tan (x) - cot (x) + C

Solution steps

\int tan^{2} (x) + cot^{2} (x) + 4 d x

Rewrite using trig identities

= \int sec^{2} (x) + cot^{2} (x) + 3 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 sec^{2} (x) d x + \int cot^{2} (x) d x + \int 3 d x

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

\int cot^{2} (x) d x = - x - cot (x)

\int 3 d x = 3 x

= tan (x) - x - cot (x) + 3 x

Simplify

= 2 x + tan (x) - cot (x)

Add a constant to the solution

= 2 x + tan (x) - cot (x) + C

\frac{x - 9 x x}{x}

y (2 x + x^{2} cot (x)) d x + x^{2} (1 + y) d y = 0

e^{3 x}

(2 x^{2} + 5) (4 x - 1)

\int \frac{5 - 50 x}{81 - 25 x ^{2}} d x

What is the integral of (tan^2(x)+cot^2(x)+4) ?
The integral of (tan^2(x)+cot^2(x)+4) is 2x+tan(x)-cot(x)+C