What is the integral of 9xsec(x)tan(x) ?

The integral of 9xsec(x)tan(x) is 9(xsec(x)-ln|tan(x)+sec(x)|)+C

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integral of 9xsec(x)tan(x)

\int 9 x sec (x) tan (x) d x

9 (x sec (x) - ln ∣ tan (x) + sec (x) ∣) + C

Solution steps

\int 9 x sec (x) tan (x) d x

Take the constant out:

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

= 9 \cdot \int x sec (x) tan (x) d x

Apply Integration By Parts

: x sec (x) - \int sec (x) d x

= 9 (x sec (x) - \int sec (x) d x)

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

= 9 (x sec (x) - ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= 9 (x sec (x) - ln ∣ tan (x) + sec (x) ∣) + C

y^{^{''}} + y = cos^{2} (x)

\int 3 v (v^{2} + 2)^{2} d v

\int (3 x 1 - 2 x^{2}) d x

4 x + 3, x^{3} + 3

\int_{0}^{\infty} \frac{1}{e ^{x} + 1} d x

What is the integral of 9xsec(x)tan(x) ?
The integral of 9xsec(x)tan(x) is 9(xsec(x)-ln|tan(x)+sec(x)|)+C