Computing Indefinite Integrals — Question 9

Question 9 - Solution

Use the trigonometric substitution $$x=\tan\theta, \qquad dx=\sec^2\theta\,d\theta.$$ Then $$1+x^2=\sec^2\theta, \qquad (1+x^2)^{3/2}=\sec^3\theta.$$

Substitute into the integral: $$\int \frac{\tan^4\theta}{\sec^3\theta}\sec^2\theta\,d\theta = \int \tan^4\theta\cos\theta\,d\theta.$$

Rewrite $\tan^4\theta$: $$\tan^4\theta = (\sec^2\theta-1)^2 = \sec^4\theta - 2\sec^2\theta + 1.$$

Thus, $$\int (\sec^3\theta - 2\sec\theta + \cos\theta)\,d\theta.$$

Integrate term by term: $$\int \sec^3\theta\,d\theta = \tfrac12\sec\theta\tan\theta + \tfrac12\ln(\sec\theta+\tan\theta),$$ $$\int \sec\theta\,d\theta = \ln(\sec\theta+\tan\theta), \qquad \int \cos\theta\,d\theta = \sin\theta.$$

Combining, $$\frac12\sec\theta\tan\theta - \frac32\ln(\sec\theta+\tan\theta) + \sin\theta + C.$$

Substitute back: $$\sec\theta=\sqrt{1+x^2}, \qquad \tan\theta=x, \qquad \sin\theta=\frac{x}{\sqrt{1+x^2}}.$$

Therefore, $$\boxed{ \frac{x}{2}\sqrt{1+x^2} - \frac{3}{2}\ln\!\left(x+\sqrt{1+x^2}\right) + \frac{x}{\sqrt{1+x^2}} + C }$$