What is the integral of 3^xsec^2(3^x) ?

The integral of 3^xsec^2(3^x) is 1/(ln(3))tan(3^x)+C

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integral of 3^xsec^2(3^x)

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

\frac{1}{ln ( 3 )} tan (3^{x}) + C

Solution steps

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

Apply u-substitution

: \int \frac{sec ^{2} ( u )}{ln ( 3 )} d u

= \int \frac{sec ^{2} ( u )}{ln ( 3 )} d u

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{ln ( 3 )} tan (u)

Substitute back

u = 3^{x}

= \frac{1}{ln ( 3 )} tan (3^{x})

Add a constant to the solution

= \frac{1}{ln ( 3 )} tan (3^{x}) + C

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What is the integral of 3^xsec^2(3^x) ?
The integral of 3^xsec^2(3^x) is 1/(ln(3))tan(3^x)+C