# How do I evaluate inttan^3(x) sec^5(x)dx?

Feb 4, 2015

The answer is: $\frac{1}{7} {\cos}^{-} 7 x - \frac{1}{5} {\cos}^{-} 5 x + c$

We can write this integral in this way:

$\int \frac{{\sin}^{3} x}{\cos} ^ 3 x \cdot \frac{1}{\cos} ^ 5 x \mathrm{dx} = \int {\sin}^{3} \frac{x}{\cos} ^ 8 x \mathrm{dx} =$

$= \int {\sin}^{2} x \sin x {\cos}^{-} 8 x \mathrm{dx} = \int \left(1 - {\cos}^{2} x\right) \sin x {\cos}^{-} 8 x \mathrm{dx} =$

$= \int \sin x {\cos}^{-} 8 x \mathrm{dx} - \int \sin x {\cos}^{-} 6 x \mathrm{dx} =$

$= - \int - \sin x {\cos}^{-} 8 x \mathrm{dx} + \int \sin x {\cos}^{-} 6 x \mathrm{dx} =$

$= - {\cos}^{- 8 + 1} \frac{x}{- 8 + 1} + {\cos}^{- 6 + 1} \frac{x}{- 6 + 1} + c = - {\cos}^{-} 7 \frac{x}{- 7} + {\cos}^{-} 5 \frac{x}{-} 5 + c =$

$= \frac{1}{7} {\cos}^{-} 7 x - \frac{1}{5} {\cos}^{-} 5 x + c$.

I have used the integral:

$\int {\left[f \left(x\right)\right]}^{n} \cdot f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$.