Recitation 4

Example 1: Evaluate \int \sec^3 x dx.

Solution: Let u=\sin x. We have \begin{aligned}I &= \int \frac{\cos x}{\cos^4 x} \\ & = \int\frac{du}{(1-u^2)^2} \\ & = \frac{1}{4}\int\left[\frac{1}{1+u}+\frac{1}{1-u}+\frac{1}{(1+u)^2}+\frac{1}{(1-u)^2}\right] \\ & = \frac{1}{4}\left[\frac{2u}{1-u^2}+\ln\frac{1+u}{1-u}\right]+C \\ & = \frac{1}{2}\frac{\sin x}{\cos^2 x}+\frac{1}{4}\ln\frac{1+\sin x}{1-\sin x}+C.\end{aligned}

Problem 2: Evaluate \int \tan x\sec^3x dx.

Hint: Let u=\sec x.

Problem 3: Evaluate \int \tan^2 xdx.

Hint: Use \tan^2 x = \sec^2 x - 1.

Problem 4: Evaluate \int\tan^4 x\sec^6 x dx.

Hint: Use u=\tan x and \sec^2 x = 1 + \tan^2 x.

Problem 5: Evaluate \int \tan^3x\sec x dx.

Hint: Let u = \sec x and use \tan^2x = \sec^2 x-1.

Problem 6: Evaluate \int \tan^5 x dx.

Hint: Let u=\sec x and use \tan^2x = \sec^2 x-1.

Problem 7: Evaluate \int \sin 8x\cos 5x dx.

Hint: Use \sin a\cos b = \frac{1}{2}(\sin(a+b)+\sin(a-b)).

Problem 8: Evaluate \int \sin 5\theta\sin\theta d\theta.

Hint: Use \sin a\sin b = \frac{1}{2}(\cos(a-b)-\cos(a+b)).

Leave a Reply

Your email address will not be published. Required fields are marked *