What is the integral of xsec^2(4x^2-5) ?

The integral of xsec^2(4x^2-5) is 1/8 tan(4x^2-5)+C

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integral of xsec^2(4x^2-5)

\int x sec^{2} (4 x^{2} - 5) d x

\frac{1}{8} tan (4 x^{2} - 5) + C

Solution steps

\int x sec^{2} (4 x^{2} - 5) d x

Apply u-substitution

: \int \frac{sec ^{2} ( 4 u - 5 )}{2} d u

= \int \frac{sec ^{2} ( 4 u - 5 )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{2} (4 u - 5) d u

Apply u-substitution

: \int sec^{2} (v) \frac{1}{4} d v

= \frac{1}{2} \cdot \int sec^{2} (v) \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int sec^{2} (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{4} tan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} tan (4 x^{2} - 5)

Simplify

\frac{1}{2} \cdot \frac{1}{4} tan (4 x^{2} - 5) : \frac{1}{8} tan (4 x^{2} - 5)

= \frac{1}{8} tan (4 x^{2} - 5)

Add a constant to the solution

= \frac{1}{8} tan (4 x^{2} - 5) + C

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What is the integral of xsec^2(4x^2-5) ?
The integral of xsec^2(4x^2-5) is 1/8 tan(4x^2-5)+C