What is the integral of 2θtan^2(θ^2) ?

The integral of 2θtan^2(θ^2) is -θ^2+tan(θ^2)+C

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integral of 2θtan^2(θ^2)

\int 2 θ tan^{2} (θ^{2}) d θ

- θ^{2} + tan (θ^{2}) + C

Solution steps

\int 2 θ tan^{2} (θ^{2}) d θ

Take the constant out:

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

= 2 \cdot \int θ tan^{2} (θ^{2}) d θ

Apply u-substitution

: \int \frac{tan ^{2} ( u )}{2} d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

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

= 2 \cdot \frac{1}{2} (- u + tan (u))

Substitute back

u = θ^{2}

= 2 \cdot \frac{1}{2} (- θ^{2} + tan (θ^{2}))

Simplify

2 \cdot \frac{1}{2} (- θ^{2} + tan (θ^{2})) : - θ^{2} + tan (θ^{2})

= - θ^{2} + tan (θ^{2})

Add a constant to the solution

= - θ^{2} + tan (θ^{2}) + C

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What is the integral of 2θtan^2(θ^2) ?
The integral of 2θtan^2(θ^2) is -θ^2+tan(θ^2)+C