What is the integral of 5^{tan(x)}sec^2(x) ?

The integral of 5^{tan(x)}sec^2(x) is (5^{tan(x)})/(ln(5))+C

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integral of 5^{tan(x)}sec^2(x)

\int 5^{t a n (x)} sec^{2} (x) d x

\frac{5 ^{t a n (x)}}{ln ( 5 )} + C

Solution steps

\int 5^{t a n (x)} sec^{2} (x) d x

Apply u-substitution

: \int 5^{u} d u

= \int 5^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{5 ^{u}}{ln ( 5 )}

Substitute back

u = tan (x)

= \frac{5 ^{t a n (x)}}{ln ( 5 )}

Add a constant to the solution

= \frac{5 ^{t a n (x)}}{ln ( 5 )} + C

\frac{d y}{d x} = \frac{4 x + x y ^{2}}{y - x ^{2} y}

e^{8 e^{x}} \cdot 8 e^{x}

\int x^{\frac{1}{2}} + cos (x) + 5 d x

\int \frac{x}{y} d x

\frac{\partial}{\partial x} (\frac{100}{4 x ^{2} + y ^{2}})

What is the integral of 5^{tan(x)}sec^2(x) ?
The integral of 5^{tan(x)}sec^2(x) is (5^{tan(x)})/(ln(5))+C