Question

Evaluate$\int {\tan }^4\mathrm{x}\mathrm{dx}$

Answer 1

$\int {\tan }^4\mathrm{x}\mathrm{d}=\int {\tan }^{2}x{\tan }^{2}x\mathrm{dx}$

$=\int ({\sec }^{2}x–1){\tan }^{2}x\mathrm{dx}$

$=\int ({\sec }^{2}x{\tan }^{2}x–{\tan }^{2}x)\mathrm{dx}$

$=\int {\sec }^{2}x{\tan }^{2}x\mathrm{dx}–\int {\tan }^{2}x\mathrm{dx}$

$=\int {\sec }^2\mathrm{x}{\tan }^2\mathrm{x}\mathrm{dx}–\int ({\sec }^2\mathrm{x}–1)\mathrm{dx}$

$=\int {\sec }^2\mathrm{x}{\tan }^2\mathrm{x}\mathrm{dx}–\int {\sec }^2\mathrm{x}\mathrm{dx}+\int 1\mathrm{dx}$

Put $\tan \mathrm{x}=\mathrm{t}$

Differentiating both the sides, we get

${\sec }^{2}x\mathrm{dx}=\mathrm{dt}$ $[\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{\mathrm{dx}}={\sec }^2\mathrm{x}]$

⇒$\int {\sec }^2\mathrm{x}{\tan }^2\mathrm{x}\mathrm{dx}–\int {\sec }^2\mathrm{x}\mathrm{dx}+\int 1\mathrm{dx}=\int {\mathrm{t}}^2\mathrm{dt}–\int \mathrm{dt}+\mathrm{x}$

$=\frac{{\mathrm{t}}^3}{3}–\mathrm{t}+\mathrm{x}+\mathrm{ca}$

$=\frac{1}{3}\left(\tan 3x\right)–\tan x+x+c$

Hence, the required answer is $\frac{1}{3}\left(\tan 3x\right)–\tan x+x+c$ .