Other Cases

There are a number of other trig functions that you might want to integrate. Here is a sketch of how to handle them. Some of these techniques are fairly advanced, so check with your instructor to see whether they are on the syllabus for your section.

• $\int \sec^n(x) \tan^m(x)\, dx$ with $n$ odd and $m$ even.
  
  First convert all of the tangents into secants, so that we have something of the form $\int \sec^n(x)\, dx$. Then do an integration by parts with $u=\sec^{n-2}(x)$ and $dv = \sec^2(x)\, dx$. After a little algebra, you get a recursive formula for $\int \sec^{n}(x)\, dx$ in terms of $\int \sec^{n-2}(x) \,dx$. Repeat the process until you get it down to $\int \sec(x)\,dx = \ln\lvert\sec(x)+\tan(x)\rvert+C$.
  
  

  Example 1: $\int \sec^3(x)\, dx$. Integrate by parts with $u=\sec(x)$, $dv=\sec^2(x)$, so $du=\sec(x)\tan(x)\, dx$ and $v=\tan(x)$:$$\int \sec^3(x) \,dx = \sec(x)\tan(x) - \int \sec(x)\tan^2(x) \,dx$$Then, since $\tan^2(x)=\sec^2(x)-1$, we can rewrite this as $$\int \sec^3(x) \,dx = \sec(x)\tan(x) + \int \sec(x) \,dx - \int \sec^3(x) \,dx.$$Adding $\int \sec^3(x)\, dx$ to each side and dividing by 2 gives$$\int \sec^3(x) \,dx = \frac{\sec(x)\tan(x) + \int \sec(x) \,dx}{2} = \frac{\sec(x)\tan(x) + \ln\lvert\sec(x)+\tan(x)\rvert}{2}+C.$$




• Integrals of the form $\int\csc^n(x)\cot^m(x)\,dx$.
  
  These follow the same strategy as $\int \sec^n(x)\tan^m(x)\, dx$, with $\csc$ replacing $\sec$ and $\cot$ replacing $\tan$. The only difference is with signs. Remember that the derivative of $\cot(x)$ is minus $\csc^2(x)$, and that the derivative of $\csc(x)$ is minus $\csc(x)\cot(x)$.



• Other products and ratios of trig functions.
  
  These can all be converted to the previous cases by writing everything in terms of sines and cosines and using the identity $\sin^2(x)+\cos^2(x)=1$.
  
  

  Example 2: Since $$\sec^2(x)\csc^2(x) = \frac{1}{\sin^2(x)\cos^2(x)} = \frac{\sin^2(x)+\cos^2(x)}{\sin^2(x)\cos^2(x)}$$ $$\qquad\qquad\qquad=\frac{1}{\cos^2(x)}+ \frac{1}{\sin^2(x)}=\sec^2(x) + \csc^2(x),$$ we have $$\int \sec^2(x)\csc^2(x)\, dx = \tan(x) - \cot(x) + C.$$

  
  
  

  Example 3: Since $$\sin(x) \tan(x) = \frac{\sin^2(x)}{\cos(x)} = \frac{1-\cos^2(x)}{\cos(x)} = \sec(x) - \cos(x),$$ we have $$\int \sin(x)\tan(x)\, dx = \int \sec(x)\, dx - \int \cos(x) \,dx = \ln\bigl\lvert\sec(x)+\tan(x)\bigr\rvert - \sin(x) +C.$$





• Products of trig functions with different frequencies, like $\sin(x)\cos(2x)$ or $\cos(2x)\cos(3x)$, can be handled either with integration-by-parts (going in circles) or by using the addition-of-angle formulas:
  \begin{eqnarray*}\sin(A)\sin(B)&=&\frac{\cos(A-B)-\cos(A+B)}{2}\cr \cos(A)\cos(B)&=&\frac{\cos(A-B)+\cos(A+B)}{2}\cr\sin(A)\cos(B)&=&\frac{\sin(A+B)+\sin(A-B)}{2}\end{eqnarray*}
  
  

  Example 4: We compute $\int \sin(x)\cos(2x)\,dx$ in two ways. By parts: $$ \int \sin(x)\cos(2x) dx \overset{\fbox{IBP: $ u\,=\,\cos(2x),\\\quad dv\,=\,\sin(x)\,dx$}}{=} -\cos(x)\cos(2x) - \int 2\cos(x)\sin(2x)\,dx $$ $$ \int \sin(x)\cos(2x) \,dx \overset{\fbox{IBP: $u\,=\,2\sin(2x),\\ \quad dv\,=\,\cos(x)\,dx$}}{=} -\cos(x)\cos(2x) - \left(2\sin(x)\sin(2x) -4\int \sin(x)\cos(2x) \,dx\right)$$ With some algebra, we find $$ -3\int\sin(x)\cos(2x)\,dx = -\cos(x)\cos(2x) - 2 \sin(x)\sin(2x) +C, $$ i.e., $$ \int\sin(x)\cos(2x)\,dx= \frac{\cos(x)\cos(2x)+2\sin(x)\sin(2x)}{3}+C$$ Or, using addition-of-angle formulas: \begin{eqnarray*}\int \sin(x) \cos(2x) dx = & \int \frac{\sin(3x)+\sin(-x)}{2} dx & \quad \hbox{(using the formula for $\sin(A)\cos(B)$)}\cr =&\int \frac{\sin(3x)-\sin(x)}{2} dx & \quad \hbox{(since $\sin(-x)=-\sin(x)$)}\cr =& \frac{-\cos(3x)}{6} + \frac{\cos(x)}{2} + C\end{eqnarray*} The two answers don't look the same, but in fact they are equal, as you can check by applying the addition-of-angle formulas to the first answer.