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

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

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integral of 1+tan^2(x)-2cos(x)

\int 1 + tan^{2} (x) - 2 cos (x) d x

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

Solution steps

\int 1 + tan^{2} (x) - 2 cos (x) d x

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= \int - 2 cos (x) + sec^{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

= - \int 2 cos (x) d x + \int sec^{2} (x) d x

\int 2 cos (x) d x = 2 sin (x)

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

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

Add a constant to the solution

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

\frac{\partial}{\partial r} (\frac{π}{3} h^{2} (3 r - h))

e^{x} y d y = (e^{- y} + e^{2 x - y}) d x

n = 1 \sum \infty n e^{5 n}

\int_{0}^{2} 10000 e^{0.04 x} d x

\int \frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} d x

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