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

The integral of θ^2sec^2(θ^3) is 1/3 tan(θ^3)+C

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integral of θ^2sec^2(θ^3)

\int θ^{2} sec^{2} (θ^{3}) d θ

\frac{1}{3} tan (θ^{3}) + C

Solution steps

\int θ^{2} sec^{2} (θ^{3}) d θ

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = θ^{3}

= \frac{1}{3} tan (θ^{3})

Add a constant to the solution

= \frac{1}{3} tan (θ^{3}) + C

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What is the integral of θ^2sec^2(θ^3) ?
The integral of θ^2sec^2(θ^3) is 1/3 tan(θ^3)+C