What is the integral of (sec(5θ)tan(5θ)+csc^2(2θ)) ?

The integral of (sec(5θ)tan(5θ)+csc^2(2θ)) is 1/5 sec(5θ)-1/2 cot(2θ)+C

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integral of (sec(5θ)tan(5θ)+csc^2(2θ))

\int (sec (5 θ) tan (5 θ) + csc^{2} (2 θ)) d θ

\frac{1}{5} sec (5 θ) - \frac{1}{2} cot (2 θ) + C

Solution steps

\int sec (5 θ) tan (5 θ) + csc^{2} (2 θ) d θ

Apply the Sum Rule:

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

= \int sec (5 θ) tan (5 θ) d θ + \int csc^{2} (2 θ) d θ

\int sec (5 θ) tan (5 θ) d θ = \frac{1}{5} sec (5 θ)

\int csc^{2} (2 θ) d θ = - \frac{1}{2} cot (2 θ)

= \frac{1}{5} sec (5 θ) - \frac{1}{2} cot (2 θ)

Add a constant to the solution

= \frac{1}{5} sec (5 θ) - \frac{1}{2} cot (2 θ) + C

\frac{\partial}{\partial y} (x y cos (yz))

\frac{d y}{d x} = \frac{x ^{2} + 3}{y}

f (x) = ln (x^{2} - 4)

\int (x - 1) e^{- x} d x

\frac{d ^{2}}{d x ^{2}} (6 x + 17)

What is the integral of (sec(5θ)tan(5θ)+csc^2(2θ)) ?
The integral of (sec(5θ)tan(5θ)+csc^2(2θ)) is 1/5 sec(5θ)-1/2 cot(2θ)+C