Integriranje trigonometrijskih funkcija

Vrijedi

$\displaystyle \int \cos ^{5}x dx$	$\displaystyle =\int \cos ^{3}x\cos ^{2}x dx=\int \cos ^{3}x\left( 1-\sin ^{2}x\right) dx$
	$\displaystyle =\int \cos ^{3}x dx-\int \cos ^{3}x\sin ^{2}x dx$
	$\displaystyle =\int \cos x\left( 1-\sin ^{2}x\right) dx-\int \cos ^{3}x\sin ^{2}x dx$
	$\displaystyle =\int \cos x dx-\int \cos x\sin ^{2}x dx-\int \cos ^{3}x\sin ^{2}x dx$
	$\displaystyle =\sin x-I_{1}-I_{2}.$

Integrale označene sa $I_{1}$ i $I_{2}$ računamo posebno koristeći jednostavne supstitucije.

$\displaystyle I_{1}$	$\displaystyle =\int \cos x\sin ^{2}x dx=\left\{ \begin{array}{c} \sin x=t \cos x dx= dt \end{array} \right\}$
	$\displaystyle =\int t^{2} dt=\frac{t^{3}}{3}+C_{2}=\frac{\sin ^{3}x}{3}+C_{1}.$
$\displaystyle I_{2}$	$\displaystyle =\int \cos ^{3}x\sin ^{2}x dx=\left\{ \begin{array}{c} \sin x=t \cos x dx= dt \end{array} \right\}$
	$\displaystyle =\int t^{2}\left( 1-t^{2}\right) dt=\int t^{2} dt-\int t^{4} dt$
	$\displaystyle =\frac{t^{3}}{3}-\frac{t^{5}}{5}+C_{2}=\frac{\sin ^{3}x}{3}-\frac{\sin ^{5}x}{5}+C_{2}.$

pa je konačno rješenje

$\displaystyle \int \cos ^{5}x dx$	$\displaystyle =\sin x-\frac{\sin ^{3}x}{3}-\frac{\sin^{3}x}{3}+ \frac{\sin ^{5}x}{5}+C$
	$\displaystyle =\sin x-\frac{2\sin ^{3}x}{3}+\frac{\sin ^{5}x}{5}+C.$

Podintegralnu funkciiju prvo raspišemo pomoću trigonometrijskih formula pretvorbe, pa vrijedi

$\displaystyle \int \cos x\cos 2x\cos 5x dx$	$\displaystyle =\frac{1}{2}\int \left( \cos x+\cos 3x\right) \cos 5x dx$
	$\displaystyle =\frac{1}{2}\int \cos x\cos 5x dx+\frac{1}{2}\int \cos 3x\cos 5x dx$
	$\displaystyle =\frac{1}{4}\int \left( \cos 4x+\cos 6x\right) dx+\frac{1}{4}\int \left( \cos 2x+\cos 8x\right) dx$
	$\displaystyle =\frac{1}{4}\bigg( \int \cos 4x dx+\int \cos 6x dx$
	$\displaystyle \quad +\int \cos 2x dx+\int \cos 8x dx\bigg)$
	$\displaystyle =\frac{1}{4}\left( \frac{\sin 4x}{4}+\frac{\sin 6x}{6}+\frac{\sin 2x}{2}+ \frac{\sin 8x}{8}\right) +C$
	$\displaystyle =\frac{\sin 2x}{8}+\frac{\sin 4x}{16}+\frac{\sin 6x}{24}+\frac{\sin 8x}{32} +C.$

Integral računamo koristeći univerzalnu trigonometrijsku supstituciju

[M2, poglavlje 1.5.1].

$\displaystyle \int \frac{ dx}{2\sin x-\cos x+5}$	$\displaystyle =\left\{ \begin{array}{cc} {\mathop{\mathrm{tg}}}\frac{x}{2}=t & ... ...,dx=\frac{2 dt}{1+t^{2}} & \cos x=\frac{1-t^{2}}{1+t^{2}} \end{array} \right\}$
	$\displaystyle =\int \frac{\frac{2 dt}{1+t^{2}}}{2\frac{2t}{1+t^{2}}-\frac{1-t^... ...{2} }+5}=\int \frac{\frac{2 dt}{1+t^{2}}}{\frac{4t-1+t^{2}+5+5t^{2}}{1+t^{2}}}$
	$\displaystyle =\int \frac{2 dt}{6t^{2}+4t+4}=\int \frac{ dt}{3\left( t^{2}+\frac{2}{3}t+ \frac{2}{3}\right) }$
	$\displaystyle =\int \frac{ dt}{\left( t+\frac{1}{3}\right) ^{2}+\frac{5}{9}}=\frac{1}{ \sqrt{5}}\mathop{\mathrm{arctg}}\nolimits \frac{3t+1}{\sqrt{5}}+C$
	$\displaystyle =\frac{1}{\sqrt{5}}\mathop{\mathrm{arctg}}\nolimits \frac{3{\mathop{\mathrm{tg}}} \frac{x}{2}+1}{\sqrt{5}}+C.$

Integrala računamo koristeći jednostavniju supstituciju:

$\displaystyle \int \frac{\cos ^{3}x+\cos ^{5}x}{\sin ^{2}x+\sin ^{4}x} dx$	$\displaystyle =\left\{ \begin{array}{c} \sin x=t \cos x dx= dt \end{array} \right\}$
	$\displaystyle =\int \frac{\cos ^{3}x\left( 1+\cos ^{2}x\right) }{\sin ^{2}x\left( 1+\sin ^{2}x\right) } dx$
	$\displaystyle =\int \frac{\cos ^{2}x\left( 1+\cos ^{2}x\right) \cos x}{\sin ^{2}x\left( 1+\sin ^{2}x\right) } dx$
	$\displaystyle =\int \frac{\left( 1-t^{2}\right) \left( 1+1-t^{2}\right) }{t^{2}... ...\left( 1-t^{2}\right) \left( 2-t^{2}\right) }{ t^{2}\left( 1+t^{2}\right) } dt$
	$\displaystyle =\int \frac{t^{4}-3t^{2}+2}{t^{4}+t} dt=\left\{ \begin{array}{c}... ...right) :\left( t^{4}+t\right) =1 \vdots ost.4t^{2}+2 \end{array} \right\}$
	$\displaystyle =\int 1 dt+\int \frac{-4t^{2}+2}{t^{2}\left( 1+t^{2}\right) } dt$
	$\displaystyle =\left\{ \begin{array}{c} \frac{-4t^{2}+2}{t^{2}\left( 1+t^{2}\ri... ...^{2}}+ \frac{Ct+D}{t^{2}+1} A=0, B=2, C=0, D=-6\end{array} \right\}$
	$\displaystyle =t+\int \frac{2}{t^{2}} dt+\int \frac{-6}{t^{2}+1} dt=t-\frac{2}{t}-6\mathop{\mathrm{arctg}}\nolimits t+C.$