What is the integral of [sin^2(x)+sec^2(5x)] ?

The integral of [sin^2(x)+sec^2(5x)] is 1/2 (x-1/2 sin(2x))+1/5 tan(5x)+C

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integral of [sin^2(x)+sec^2(5x)]

\int [sin^{2} (x) + sec^{2} (5 x)] d x

\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{5} tan (5 x) + C

Solution steps

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

Apply the Sum Rule:

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

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

\int sin^{2} (x) d x = \frac{1}{2} (x - \frac{1}{2} sin (2 x))

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

= \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{5} tan (5 x)

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{5} tan (5 x) + C

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What is the integral of [sin^2(x)+sec^2(5x)] ?
The integral of [sin^2(x)+sec^2(5x)] is 1/2 (x-1/2 sin(2x))+1/5 tan(5x)+C