Question

Find the
∫tan3xsec3xdx

Answer 1

Answer on Question #45970 – Math – Integral Calculus

Question.

Find the $\int \tan 3x \sec 3x \, dx$.

Solution.

$$\tan \theta = \frac{\sin \theta}{\cos \theta};$$ $$\sec \theta = \frac{1}{\cos \theta}$$ $$\int \tan 3x \cdot \sec 3x \, dx = \int \frac{\sin 3x}{\cos 3x} \cdot \frac{dx}{\cos 3x} = \int \frac{\sin 3x}{(\cos 3x)^2} \, dx = \{y = 3x\} = \frac{1}{3} \int \frac{\sin y}{(\cos y)^2} \, dy = \\ = -\frac{1}{3} \int \frac{d(\cos y)}{(\cos y)^2} = \{z = \cos y\} = -\frac{1}{3} \int \frac{dz}{z^2} = \frac{1}{3z} + C = \{z = \cos y\} = \frac{1}{3\cos y} + C = \\ = \{y = 3x\} = \\$$

$= \frac{1}{3\cos 3x} + C$, where $C$ is an arbitrary real constant.

Answer.

$$\int \tan 3x \cdot \sec 3x \, dx = \frac{1}{3\cos 3x} + C$$

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