Question

Find the
∫tan^3xsec^3xdx

tan^2x+1
cot^2x+1
sec^2x+1
sec^2x

Answer 1

Answer to Question #46133 – Math – Integral Calculus

Find the /tan^3xsec^3xdx

$$\begin{array}{l} \tan^2x + 1 \\ \cot^2x + 1 \\ \sec^2x + 1 \\ \sec^2x \end{array}$$

Solution:

$$\begin{array}{l} \int \tan^3 x * \sec^3 x \, dx = \int \frac{\sin^3 x}{\cos^3 x} * \frac{1}{\cos^3 x} \, dx = \\ = - \int \frac{\sin^2 x}{\cos^6 x} \, d\cos x = \int \frac{\cos^2 x - 1}{\cos^6 x} \, d\cos x = \int \left( \frac{1}{\cos^4 x} - \frac{1}{\cos^6 x} \right) d\cos x = \\ = \frac{1}{5 * \cos^5 x} - \frac{1}{3 * \cos^3 x} + C \end{array}$$

where $C$ is an arbitrary constant.

Answer:

$$\int \tan^3 x * \sec^3 x \, dx = \frac{1}{5 * \cos^5 x} - \frac{1}{3 * \cos^3 x} + C$$

P. S.: maybe you make some mistake in task /tan^3xsec^3xdx.

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