Integration Techniques, multiple pages

7. Trigonometric and hyperbolic integrals

Since this site is heavily invested in trigonmetry, we will not come back to formulas we have already seen in the trigometry field.

We’ll focus on some techniques used to integrate trigonometric and hyperbolic functions.

We have already used some of these.

Cases involving $\sin x$ and $\cos x$

We use the following:

$\tan \frac{x}{2}=t$

We can easily show that:

$\sin x=\frac{2t}{1+t^{2}}$

$\cos x=\frac{1-t^{2}}{1+t^{2}}$

$x=2 \arctan t$

$dx=\frac{2dt}{1+t^{2}}$

And with are back to normal rational integration.

Cases of form $\int \sin^{m} x \cos^{n} x\, dx$

-If $m$ is ODD, use $u=\cos x$ .

-If $n$ is ODD, use $u=\sin x$ .

Cases of form $\int \tan^{m} x \sec^{n} x\, dx$

-If $m$ is ODD, use $u=\sec x$ .

-If $n$ is EVEN, use $u=\tan x$ .

Problem 39

Evaluate: $\int \sin^{3} x \, dx$

we can see that:

$\sin^{3} x=\sin^{2} x \sin x =(1-\cos^{2} x) \sin x$

Let $u=\cos x \Rightarrow du=-\sin x \, dx$

${\displaystyle \int \sin^{3} x \, dx}= {\displaystyle -\int (1-u^{2})\,du }\\={\displaystyle -\int du+\int u^{2}\,du }\\=-u+\frac{u^{3}}{3}+C=-\cos x+\frac{1}{3} \cos^{3}x+C$

Finally:

$\int \sin^{3} x \, dx}={\displaystyle -\cos x+\frac{1}{3} \cos^{3}x+C$

Problem 40

Evaluate: $\int x \arcsin x \, dx$

We’ll use integration by parts:

Let $u=\arcsin x \Rightarrow du=\frac{dx}{\sqrt{1-x^{2}}}$

$dv=xdx \Rightarrow v=\frac{1}{2}x^{2}$

We get:

$\int x \arcsin x \, dx }={\displaystyle \frac{1}{2}x^{2} \arcsin x-\frac{1}{2}\int \frac{x^{2} \,dx}{\sqrt{1-x^{2}}}$

Let:

$I= -\frac{1}{2}\int \frac{x^{2} \,dx}{\sqrt{1-x^{2}}}$

We use trigonometric substitution:

Let $\sin \theta=x \Rightarrow dx=\cos \theta \, d\theta$

$\cos \theta=\sqrt{1-x^{2}}$

$\theta=\arcsin x$

(1) $\begin{equation*} \begin{split} I&= {\displaystyle -\frac{1}{2}\int \frac{x^{2} \,dx}{\sqrt{1-x^{2}}} }\\ &={\displaystyle -\frac{1}{2}\int \frac{\sin^{2} \theta \cos \theta \, d\theta}{\cos \theta} }\\ &= {\displaystyle -\frac{1}{2}\int \sin^{2} \theta \, d\theta }\\ &={\displaystyle -\frac{1}{4}\int (1-\cos 2\theta) \, d\theta }\\ &={\displaystyle -\frac{1}{4} \theta+\frac{1}{8}\sin 2\theta +C}\\ &={\displaystyle -\frac{1}{4} \theta +\frac{1}{4}\sin \theta \cos \theta +C}\\ &= {\displaystyle -\frac{1}{4} \arcsin x +\frac{1}{4}x \sqrt{1-x^{2}}+C} \end{split} \end{equation*}$

Back to the original:

$\int x \arcsin x \, dx }={\displaystyle \frac{1}{2}x^{2} \arcsin x-\frac{1}{4} \arcsin x +\frac{1}{4}x \sqrt{1-x^{2}}+C$

$\int x \arcsin x \, dx }={\displaystyle \frac{1}{4}(2x^{2}-1) \arcsin x+\frac{1}{4}x \sqrt{1-x^{2}}+C$

Finally:

$\int x \arcsin x \, dx }={\displaystyle \frac{1}{4}(2x^{2}-1) \arcsin x+\frac{1}{4}x \sqrt{1-x^{2}}+C$

Problem 41

Evaluate: $\int \cos^{6} x \, dx$

$\int \cos^{6} x \, dx }={\displaystyle \int (\cos^{2} x)^{3} \, dx$

$\int (\cos^{2} x)^{3} \, dx }={\displaystyle \int (\frac{1+\cos 2x }{2})^{3} \, dx$

$\int (\frac{1+\cos 2x }{2})^{3} \, dx }={\displaystyle \frac{1}{8} \int (1+\cos 2x)^{3} \, dx$

Let’s make this integrand easy to compute:

(2) $\begin{equation*} \begin{split} \displaystyle (1+\cos 2x)^{3} }&= {\displaystyle 1+ 3\cos 2x + 3 \cos^{2} 2x+ \cos^{3} 2x}\\ &={\displaystyle 1+ 3\cos 2x+ 3 \frac{1+\cos 4x}{2}+ (1-\sin^{2} 2x)\cos 2x }\\ &={\displaystyle 1+ 3\cos 2x+ \frac{3}{2}+ \frac{3}{2}\cos 4x+\cos 2x- \sin^{2} 2x \cos 2x }\\ &={\displaystyle \frac{5}{2}+ 3\cos 2x+ \frac{3}{2}\cos 4x+\cos 2x- \sin^{2} 2x \cos 2x } \end{split} \end{equation*}$

Back to the Equation:

(3) $\begin{equation*} \begin{split} \displaystyle \int \cos^{6} x \, dx }&={\displaystyle \frac{1}{8} \int (1+\cos 2x)^{3} \, dx }\\ &={\displaystyle \frac{1}{8} \cdot \frac{5}{2} \int dx+\frac{1}{8} \cdot 4 \int \cos 2x \,dx+ \frac{1}{8} \cdot \frac{3}{2} \int \cos 4x \,dx- \frac{1}{16} \int 2\sin^{2} 2x \cos 2x \,dx }\\ &={\displaystyle \frac{5}{16} x+ \frac{4}{16} \sin 2x+ \frac{3}{64} \sin 4x-\frac{1}{48}\sin^{3} 2x+ C}\\ &={\displaystyle \frac{1}{16} \left [ 5x+ 4\sin 2x+ \frac{3}{4} \sin 4x-\frac{1}{3}\sin^{3} 2x \right ]+ C} \end{split} \end{equation*}$

Finally:

$\int \cos^{6} x \, dx }={\displaystyle \frac{1}{16} \left [ 5x+ 4\sin 2x+ \frac{3}{4} \sin 4x-\frac{1}{3}\sin^{3} 2x \right ]+ C$

Problem 42

Evaluate: $\int \frac{\cos\theta +3}{2-\cos\theta}\mathrm{d} \theta$

$\int \frac{\cos\theta +3}{2-\cos\theta} \mathrm{d} \theta=\int \frac{\cos\theta-2 +5}{2-\cos\theta}\mathrm{d} \theta$
$\int \frac{\cos\theta-2 +5}{2-\cos\theta}\mathrm{d} \theta=-\int \frac{-\cos\theta+2}{2-\cos\theta}\mathrm{d} \theta+5\int \frac{\mathrm{d} \theta}{2-\cos\theta}$
$-\int \frac{-\cos\theta+2}{2-\cos\theta}\mathrm{d} \theta+5\int \frac{\mathrm{d} \theta}{2-\cos\theta}=-\int \mathrm{d} \theta+5\int \frac{\mathrm{d} \theta}{2-\cos\theta}$

We take:
$I=5\int \frac{\mathrm{d} \theta}{2-\cos\theta}$
Let:
$\tan \frac{\theta}{2}=x$
$\theta=2\tan^{-1}x$
$\mathrm{d} \theta=\frac{2 \mathrm{d} x}{1+x^2}$

$\cos \theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}$
$\cos \theta=\frac{1-x^2}{1+x^2}$
Now
$I=5\int \frac{\mathrm{d} \theta}{2-\cos\theta}=2\times 5\int \frac{\mathrm{d} x}{(1+x^2)\cdot(2-\frac{1-x^2}{1+x^2})}$
$I=10\int \frac{\mathrm{d} x}{2+2x^2-1+x^2}$
$I=10\int \frac{\mathrm{d} x}{3x^2+1}$
$I=\frac{10}{3}\int \frac{\mathrm{d} x}{x^2+\frac{1}{3}}$
$I=\frac{10}{3}\int \frac{\mathrm{d} x}{x^2+(\frac{1}{\sqrt{3}})^2}$
$I=\frac{10}{3}\int \frac{\mathrm{d} x}{x^2+(\frac{\sqrt{3}}{3})^2}$
$I=\frac{10}{3}\cdot \frac{3}{\sqrt{3}}\tan^{-1}\frac{3x}{\\sqrt{3}}+C_1$
Back to the angle
$I=\frac{3}{\sqrt{3}}\tan^{-1}(\frac{3\tan \frac{\theta}{2}}{\sqrt{3}})+C_1$
$\int \frac{\cos\theta +3}{2-\cos\theta}\mathrm{d} \theta=\frac{3}{\sqrt{3}}\tan^{-1}(\frac{3\tan \frac{\theta}{2}}{\sqrt{3}})-\theta+C$