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

The integral of xsec^2(5x) is 1/2 x^2sec^2(5x)-1/50 (25x^2tan^2(5x)-10xtan(5x)-2ln|cos(5x)|+25x^2)+C

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integral of xsec^2(5x)

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

\frac{1}{2} x^{2} sec^{2} (5 x) - \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2}) + C

Solution steps

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

Rewrite using trig identities

= \int x (1 + tan^{2} (5 x)) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} sec^{2} (5 x) - \int 5 x^{2} sec^{2} (5 x) tan (5 x) d x

= \frac{1}{2} x^{2} sec^{2} (5 x) - \int 5 x^{2} sec^{2} (5 x) tan (5 x) d x

\int 5 x^{2} sec^{2} (5 x) tan (5 x) d x = \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2})

= \frac{1}{2} x^{2} sec^{2} (5 x) - \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2})

Add a constant to the solution

= \frac{1}{2} x^{2} sec^{2} (5 x) - \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2}) + C

y^{^{''}} + 4 y = - 2

\int cos (φ) d φ

y^{'} = 0.027 y (1 - \frac{y}{1527})

ln (3) + e^{x^{e}}

y^{^{''}} - 5 y^{^{'}} = 10 x + 3

What is the integral of xsec^2(5x) ?
The integral of xsec^2(5x) is 1/2 x^2sec^2(5x)-1/50 (25x^2tan^2(5x)-10xtan(5x)-2ln|cos(5x)|+25x^2)+C