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

The integral of sec^2(3x)y is (y^2sec^2(3x))/2+C

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integral of sec^2(3x)y

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

\frac{y ^{2} sec ^{2} ( 3 x )}{2} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{y ^{2}}{2}

= sec^{2} (3 x) \frac{y ^{2}}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{y ^{2} sec ^{2} ( 3 x )}{2}

Add a constant to the solution

= \frac{y ^{2} sec ^{2} ( 3 x )}{2} + C

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What is the integral of sec^2(3x)y ?
The integral of sec^2(3x)y is (y^2sec^2(3x))/2+C