Question

hw part 1: 7.2 trigonometric integrals
score: 3.5/8 answered: 5/8
progress saved done
question 6
0/1 pt 3 98 details
evaluate ∫tan⁴x sec⁴x dx

Explanation:

Step1: Rewrite the integrand

We know that $\sec^{2}x = 1+\tan^{2}x$. So, $\int\tan^{4}x\sec^{4}x dx=\int\tan^{4}x\sec^{2}x\sec^{2}x dx=\int\tan^{4}x(1 + \tan^{2}x)\sec^{2}x dx$.

Step2: Use substitution

Let $u=\tan x$, then $du=\sec^{2}x dx$. The integral becomes $\int u^{4}(1 + u^{2})du=\int(u^{4}+u^{6})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int(u^{4}+u^{6})du=\frac{u^{5}}{5}+\frac{u^{7}}{7}+C$.

Step4: Substitute back

Substituting $u = \tan x$ back, we get $\frac{\tan^{5}x}{5}+\frac{\tan^{7}x}{7}+C$.

Answer:

$\frac{\tan^{5}x}{5}+\frac{\tan^{7}x}{7}+C$