What is the integral of 7(tan^2(x)+tan^4(x)) ?

The integral of 7(tan^2(x)+tan^4(x)) is 7/3 tan^3(x)+C

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integral of 7(tan^2(x)+tan^4(x))

\int 7 (tan^{2} (x) + tan^{4} (x)) d x

\frac{7}{3} tan^{3} (x) + C

Solution steps

\int 7 (tan^{2} (x) + tan^{4} (x)) d x

Take the constant out:

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

= 7 \cdot \int tan^{2} (x) + tan^{4} (x) d x

Rewrite using trig identities

= 7 \cdot \int - 1 + sec^{2} (x) + tan^{4} (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

= 7 (- \int 1 d x + \int sec^{2} (x) d x + \int tan^{4} (x) d x)

\int 1 d x = x

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

\int tan^{4} (x) d x = \frac{1}{3} tan^{3} (x) + x - tan (x)

= 7 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x))

Simplify

7 (- x + tan (x) + \frac{1}{3} tan^{3} (x) + x - tan (x)) : \frac{7}{3} tan^{3} (x)

= \frac{7}{3} tan^{3} (x)

Add a constant to the solution

= \frac{7}{3} tan^{3} (x) + C

y^{^{''}} + y = t, y (0) = 1, y^{^{'}} (0) = - 1

\frac{\partial}{\partial y} (\frac{x}{x ^{7} - y ^{3}})

\frac{d}{d x} (\frac{3 x ^{2} + 1}{x + 2})

\int \frac{x + 2}{( 3 + 2 x - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (a e^{b x})

What is the integral of 7(tan^2(x)+tan^4(x)) ?
The integral of 7(tan^2(x)+tan^4(x)) is 7/3 tan^3(x)+C