Calculus

Other Trigonometric Integrals

Often we have to apply trigonometric identities before integrating. This is required when integrating the form where and n are integers, $$\int \sin^m x\cos^n x\,dx$$. For this form of integrals, there are three cases to consider.

Case 1. m is odd.

1. Write $$\sin^m x$$ as $$\sin^{m-1} x\sin x$$
2. Use $$\cos^2 x+\sin^2 x=1$$ to write the even power of sines in terms of cosines
3. Use the substitution $$u=\cos x$$

Case 2. is odd.

1. Write $$\cos^n x$$ as $$\cos^{n-1} x\cos x$$
2. Use $$\cos^2 x+\sin^2 x=1$$ to write the even power of cosines in terms of sines
3. Use the substitution $$u=\sin x$$

Case 3. Both m and are even

Use $$2\cos^2 x=1+\cos 2x$$ and $$2\sin^2 x=1-\cos 2x$$ to halve the even powers.

Example 1: Integrate $$\int \sin^3 x\cos^6 x\,dx$$

Solution: We can use the steps outlined in Case 1 since $$m=3$$.

$\int \sin^3 x\cos^6 x\,dx=\int \sin^2 x\sin x\cos^6 x\,dx$

Now we substitute $$\sin^2 x=1-\cos^2 x$$

$\int \sin^2 x\sin x\cos^6 x\,dx=\int \left(1-\cos^2 x\right)\sin x\cos^6 x\,dx=\int \sin x\cos^6 x\,dx-\int \sin x\cos^8 x\,dx$

Now let $$u=\cos x$$, then $$du=-\sin x\,dx$$

\begin{align} \int \sin x\cos^6 x\,dx-\int \sin x\cos^8 x\,dx &= \int u^6\,du-\int u^8\,du \\ &= \frac{u^7}{7} -\frac{u^9}{9}+C \\ &= \frac{\cos^7 x}{7} -\frac{\cos^9 x}{9}+C \end{align}

Example 2: Integrate $$\int \cos^2 2x\,dx$$

Solution: For this integral, we will follow the steps for Case 3 and use our double angle identities

\begin{align} \int \cos^2 2x\,dx&=\int \frac{1+\cos 4x}{2}dx\\&=\int \frac{1}{2} + \int \frac{\cos 4x}{2} \\ &=\frac{x}{2}+\frac{\sin 4x}{8}+C\end{align}

For the integrals of the form $\int \tan^m x\sec^nx\,dx$ or $\int \cot^m x\csc^nx\,dx$

Case 1. and are both odd

1. Set aside a factor $$\sec x\tan x$$ or $$\csc x\cot x$$
2. Use equations $$1+\tan^2x=\sec^2x$$ to write the even power of tangents in terms of secants, or $$1+\cot^2x=\csc^2x$$ to write the even power of cotangents in terms of cosecants
3. Use the substitution $$u=\sec x$$ or $$u=\csc x$$

Case 2. n is even

1. If possible, set aside a factor $$\sec^2 x$$ or $$\csc^2 x$$. If not possible, use $$1+\tan^2x=\sec^2x$$ to change a factor $$tan^2x$$ into a term with  $$\sec^2 x$$, or $$1+\cot^2x=\csc^2x$$ to change a factor $$cot^2x$$ into a term with $$csc^2x$$. Repeat if necessary.
2. Use $$1+\tan^2x=\sec^2x$$ to write the even power of secants in terms of tangents, or $$1+\cot^2x=\csc^2x$$ to write the even power of cosecants in terms of cotangents
3. Use the substitution $$u=\tan x$$ or $$u=\cot x$$

Case 3. is even, n is odd

1. Use $$1+\tan^2x=\sec^2x$$ to write the even power of tangents in terms of secants, or $$1+\cot^2x=\csc^2x$$ to write the even power of cotangents in terms of cosecants
2. Integrate the odd power of secants or cosecants by parts

Example 3: Integrate $$\int_{0}^{\pi/4} \tan x\sec^4 x\,dx$$

Solution: Let's set aside a $$\sec^2 x$$ and use the identity $$\sec^2 x = 1+\tan^2 x$$

\begin{align} \int_{0}^{\pi/4} \tan x\sec^4 x\,dx&=\int_{0}^{\pi/4} \tan x\sec^2 x\left(1+\tan^2\right)dx \\ &= \int_{0}^{\pi/4} \tan x\sec^2 x +\int_{0}^{\pi/4} \tan^3 x\sec^2 x \end{align}

Let $$u=\tan x$$, then $$du=\sec^2 x\,dx$$

\begin{align} \int_{0}^{\pi/4} \tan x\sec^2 x +\int_{0}^{\pi/4} \tan^3 x\sec^2 x &= \int_{x=0}^{x=\pi/4} u\,du +\int_{0}^{\pi/4} u^3 du \\ &= \left(\frac{u^2}{2} + \frac{u^4}{4}\right)_{x=0}^{x=\pi/4} \\ &=\left(\frac{\tan^2 x}{2} + \frac{\tan^4 x}{4}\right)_{0}^{\pi/4} \\&=\frac{3}{4}\end{align}

Alternative Solution: We can set aside $$\sec x\tan x$$ instead and solve this another way.

$\int_{0}^{\pi/4} \tan x\sec^4 x\,dx=\int_{0}^{\pi/4} \left(\tan x\sec x\right)\sec^3 x\,dx$

Let $$u=\sec x$$, then $$du=\sec x\tan x\,dx$$. When $$x=0$$, $$u=1$$ and when $$x=\pi/4$$, $$u=\sqrt{2}$$

\begin{align} \int_{0}^{\pi/4} \left(\tan x\sec x\right)\sec^3 x\,dx &= \int_{1}^{\sqrt{2}} u^3 du \\ &= \left(\frac{u^4}{4}\right)_{1}^{\sqrt{2}} \\&= \frac{3}{4} \end{align}

Example 4: Integrate $$\int \frac{1-\cot x}{\sin^4 x}dx$$

Solution: In some cases, you will have to use trigonometric identities and simplify before integrating

\begin{align} \int \frac{1-\cot x}{\sin^4 x}dx&=\int \frac{1}{\sin^4 x}dx- \int \frac{\cot x}{\sin^4 x}dx \\ &= \int \sec^4 x\,dx - \int \frac{\frac{\cos x}{\sin x}}{\sin^4 x}dx \\ &=  \int \left(1+\tan^2 x\right)\sec^2 x\,dx -  \int \frac{\cos x}{\sin^5 x}dx\\ &= \int\sec^2x\,dx+\int \tan^2x\sec^2 x\,dx -  \int \frac{\cos x}{\sin^5 x}dx \end{align}

For the second integral, let $$u=\tan x$$, then $$du=\sec^2 x\,dx$$. For the third integral, let $$v=\sin x$$, then $$dv=\cos x\,dx$$.

\begin{align} \int\sec^2x\,dx+\int \tan^2x\sec^2 x\,dx -  \int \frac{\cos x}{\sin^5 x}dx &= \int\sec^2x\,dx+\int u^2\,du-\int v^{-5}\,dv \\&=\tan x+\frac{u^3}{3}-\frac{v^{-4}}{-4} +C\\&=\tan x+\frac{\tan^3x}{3}+\frac{\sin^{-4}x}{4}+C  \end{align}

Example 5: Integrate $$\int\frac{\sec^2t\tan t}{4+\sec^2t}dt$$

Solution: Sometimes you have to apply multiple techniques. In this case, the denominator is of the power -1. This suggests a logarithmic integral. So we let $$u=4+\sec^2t$$, then $$du=2\sec t\left(\sec t\tan t\right)\,dt$$

\begin{align} \int\frac{\sec^2t\tan t}{4+\sec^2t}dt &= \frac{1}{2}\int\frac{du}{u} \\&= \frac{1}{2} \ln|u|+C \\&= \frac{1}{2} \ln|4+\sec^2t|+C\end{align}