Integration of Trigonometric Functions

We first rewrite the integral in the form: \[ \int \sqrt[3]{\cos x}\cdot \sin^3x\,dx=\int \left(\cos x\right)^{\frac13} \sin^2x \cdot \sin x\,dx.\] We now use the substitution \(u=\cos x\). Then we have \(du=-\sin x\,dx\) and the integral becomes: \begin{eqnarray*} \int \sqrt[3]{\cos x}\cdot \sin^3x\,dx&=&-\int u^{\frac13}\cdot (1-u^2)\,du=-\int u^{\frac13}\,du+\int u^{\frac73}\,du\\&=&-\frac34u^{\frac43}+\frac3{10}u^{\frac{10}3}+C\\&=& -\frac34 \cos^{\frac43}x+\frac{3}{10}\cos^{\frac{10}3}x+C. \end{eqnarray*}

We first rewrite the integral in the form: \[ \int \frac1{\cos x}\,dx=\int \frac{\cos x}{\cos^2x}\,dx.\] We now use the substitution \(u=\sin x\). Then we have \(du=\cos x\,dx\) and the integral becomes: \begin{eqnarray*} \int \frac1{\cos x}\,dx&=& \int\frac{du}{1-u^2}=\int\frac1{(1-u)(1+u)}\,du\\ &=&\frac12\int \frac1{1-u}\,du+\frac12\int\frac1{1+u}\,du=-\frac12\ln\left|1-u\right|+\frac12\ln\left|1+u\right|+C\\ &=&\frac12\ln\left|\frac{1+u}{1-u}\right|+C=\frac12\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C =\frac12\ln\left|\frac{1+\sin x}{1-\sin x}\cdot\frac{1+\sin x}{1+\sin x}\right|+C\\&=& \frac12\ln\left|\frac{\left(1+\sin x\right)^2}{\cos^2x}\right|+C=\ln\left|\frac{1+\sin x}{\cos x}\right|+C\\ &=&\ln\left|\sec x+\tan x\right|+C. \end{eqnarray*}

Using the formulas \(\cos^2\theta=\frac{1+\cos(2\theta)}2\) and \(\sin^2\theta=\frac{1-\cos(2\theta)}2\) we transform the integral in the equivalent form: \[ \int \cos^4x\,\sin^2 x\,dx= \int \left(\frac{1+\cos(2x)}2\right)^2\cdot\frac{1-\cos(2x)}2\,dx=\frac18\int \left(1+\cos(2x)\right)^2 \left(1-\cos(2x)\right)\,dx.\] We now use the substitution \(u=2x\). Then \(dx=\frac12du\), hence \begin{eqnarray*} \int \cos^4x\,\sin^2 x\,dx&=&\frac1{16}\int (1+\cos u)^2(1-\cos u)\,du=\frac1{16}\int\left(1+2\cos u+\cos^2 u\right)\left(1-\cos u\right)\,du\\ &=&\frac1{16}\left(\int 1\,du+\int \cos u\,du-\int \cos^2 u\,du -\int \cos^3 u\,du\right). \end{eqnarray*} The first two integrals are easy, and they are equal to \(u+C\) and \(\sin u+C\), respectively. The last integral is of the form \(\displaystyle \int \sin^mx\,\cos^n x\,dx\) where \(m=0\) and \(n=3\), which due to the fact that \(n\) is odd can be solved using the substitution \(v=\sin u\): \[\int \cos^3u\,du=\int \cos^2u\cdot \cos u\,du=\int (1-v^2)\,dv=v-\frac13v^3+C=\sin u-\frac13\sin^3u+C.\] The third integral \(\int \cos^2u\,du\) is solved in the following way: \[\int\cos^2u\,du=\int\frac{1+\cos(2u)}2\,du=\frac12u+\frac14\sin(2u).\] Therefore, \begin{eqnarray*} \int \cos^4x\,\sin^2 x\,dx&=& \frac1{16}\left(u+\sin u-\frac12u-\frac14\sin(2u)-\sin u+\frac13\sin^3u\right)+C\\ &=&\frac1{16}\left(x-\frac14\sin(4x)+\frac13\sin^3(2x)\right)+C \end{eqnarray*}

We divide both numerator and denominator by \(\cos^3 x\). The integral becomes \[ \int \frac{2\sin^3x+\sin^2x\cos x-4\sin x\cos^2x+3\cos^3x}{\left(\sin^2x-\sin x\cos x-2\cos^2x\right)\cos x}\,dx=\int\frac{2\tan^3x +\tan^2 x-4\tan x+3}{\tan^2x -\tan x-2}\,dx.\] Using the substitution \(\tan x=t\) we have \(dx=\frac1{1+t^2}\,dt\) and the integral is further equal to: \[\int \frac{2\sin^3x+\sin^2x\cos x-4\sin x\cos^2x+3\cos^3x}{\left(\sin^2x-\sin x\cos x-2\cos^2x\right)\cos x}\,dx=\int\frac{2t^3+t^2-4t+3}{(t-2)(t+1)(t^2+1)}\,dt.\] Using the method of partial fractions we find the following representation of the integrand: \[ \frac{2t^3+t^2-4t+3}{(t-2)(t+1)(t^2+1)}=\frac1{t-2}-\frac1{t+1}+\frac{2t}{t^2+1}.\] Therefore we have: \begin{eqnarray*} \int \frac{2\sin^3x+\sin^2x\cos x-4\sin x\cos^2x+3\cos^3x}{\left(\sin^2x-\sin x\cos x-2\cos^2x\right)\cos x}\,dx&=& \int\frac1{t-2}\,dt-\int\frac{1}{t+1}\,dt+\int \frac{2t}{t^2+1}\,dt\\ &=&\ln\left|t-2\right|-\ln\left|t+1\right|+\ln\left|t^2+1\right|+C\\ &=&\ln\left|\tan x-2\right|-\ln\left|\tan x+1\right|+\ln\left|\frac1{\cos^2x}\right|+C\\ &=&\ln\left|\tan x-2\right|-\ln\left|\tan x+1\right|-2\ln\left|\cos x\right|+C. \end{eqnarray*}

The substitution \(u=\tan\frac{x}2\) transforms the integral into: \begin{eqnarray*} \int \frac1{1+\sin x}\,dx&=&\int\frac{\frac{2\,du}{1+u^2}}{1+\frac{2u}{1+u^2}} =\int\frac{2\,du}{1+u^2+2u}\\ &=&\int\frac{2\,du}{(1+u)^2}=-\frac2{1+u}+C=-\frac2{ 1+\tan\frac{x}2}+C. \end{eqnarray*}

We will use the substitution \(\sin x=u\). Then we have \(du=\cos x\,dx\) and the integral becomes: \[ \int \sin^{10}x\cos x \,dx=\int u^{10}\,du=\frac1{10+1}u^{10+1}+C=\frac{\sin^{10+1}x}{10+1}+C.\]

We will use the substitution \(u=\cos x\). Then \(du=-\sin x\,dx\). The integral is equivalent to: \begin{eqnarray*}\int \sin^3 x\cos^{6}x\,dx&=&\int \left(1-\cos^2x\right)\cos^{6}x \cdot \sin x\,dx=-\int \left(1-u^2\right)u^{6}\,du \\&=& -\frac{u^{6+1}}{6+1}+\frac{u^{6+3}}{6+3}+C=-\frac{\cos^9x}9+\frac{\cos^{11}x}{11}+C. \end{eqnarray*}

The original integral can be rewritten in the form: \[\int_0^{\frac{\pi}4}\frac{ 3\sin x +2 \cos x}{ 4\sin x+ 3\cos x} \,dx=\int_0^{\frac{\pi}4}\frac{3\tan x +2}{4\tan x+3}\,dx.\] We will use the substitution \(u=\tan x\). Then we have \(x=\tan^{-1}u\) and \(\displaystyle dx=\frac{1}{1+u^2}\,du\). The bounds for \(u\) are \(0\leq u\leq 1\) and the integral becomes \[\int_0^{\frac{\pi}4}\frac{ 3\sin x +2 \cos x}{ 4\sin x+ 3\cos x} \,dx=\int_0^1\frac{ 3 u +2}{(4u+3)\left(u^2+1\right)}\,du.\] The integral is reduced to the integral of a rational function. We need to find constants \(\alpha\), \(\beta\), and \(\gamma\) such that: \[\frac{ 3 u +2}{(4u+3)\left(u^2+1\right)}=\frac{\alpha}{4u+3}+\frac{\beta u+\gamma}{4u+3}.\] Using methods described in the section _link_i1|Integration of rational functions|676_link_ we find that \(\alpha=\frac{-4}{25}\), \(\beta=\frac{1}{25}\), and \(\gamma=\frac{18}{25}\). Thus the integral reduces to: \begin{eqnarray*}\int_0^1\frac{ 3 u +2}{(4u+3)\left(u^2+1\right)}\,du&=&\frac{-4}{25}\cdot \int_0^1 \frac1{4u+3}\,du +\frac{1}{25}\cdot \int_0^1\frac{u\,du}{u^2+1} +\frac{18}{25}\cdot \int_0^1\frac{du}{u^2+1}\\ &=&\left.\frac{-4}{25}\cdot \frac1{4}\cdot \ln\left|4u+3\right|\right|_{u=0}^{u=1} +\frac{1}{25}\cdot\left.\frac12\cdot \ln\left|u^2+1\right|\right|_{u=0}^{u=1} +\frac{18}{25} \cdot \left. \tan^{-1}u\right|_{u=0}^{u=1}\\ &=&-\frac{1}{25}\ln\frac{7}{3\sqrt2}+\frac{9}{50}\pi. \end{eqnarray*}

We will use the substitution \(u=\tan\frac{x}2\). Then we have: \begin{eqnarray*} u&=&\tan\frac{x}2\\ dx&=&\frac{2\,du}{u^2+1}\\ \sin x&=&\frac{2u}{u^2+1}\\ \cos x&=&\frac{1-u^2}{1+u^2}\\ 0\leq &u&\leq 1.\end{eqnarray*} The bounds for \(u\) have been obtained in the following way. The bounds for \(x\) were \(0\) and \(\frac{\pi}2\). The corresponding bounds for \(\frac{x}2\) are \(0\) and \(\frac{\pi}4\), hence the bounds for \(u\) are \(\tan 0=0\) and \(\tan\frac{\pi}4=1\). The integral now becomes: \begin{eqnarray*} \int_0^{\frac{\pi}2}\frac1{6 +10\cos x}\,dx&=&\int_0^1\frac{\frac{2\,du}{1+u^2}}{6+10\frac{1-u^2}{1+u^2}}=\int_0^1\frac{2\,du}{(6+10)-(10 -6)u^2}=\int_0^1 \frac{2\,du}{16 -4u^2}\\ &=&\int_0^1\frac{2\,du}{(4+2u)(4-2u)}= \int_0^1\left(\frac1{4\left(4+2u\right)}+\frac1{4\left(4-2u\right)}\right)\,du\\ &=&\frac1{4}\int_0^1\frac1{4+2u}\,du+\frac1{4}\int_0^1\frac1{4-2u}\,du\\ &=&\left.\frac1{8}\ln\left|\frac{4+2u}{4-2u}\right|\right|_0^1=\frac1{8}\ln\left|\frac{4+2}{4-2}\right|\\ &=&\frac1{8}\ln3. \end{eqnarray*}

We will use the substitution \(u=\tan\frac{x}2\). Then we have: \begin{eqnarray*} u&=&\tan\frac{x}2\\ dx&=&\frac{2\,du}{u^2+1}\\ \sin x&=&\frac{2u}{u^2+1}\\ \cos x&=&\frac{1-u^2}{1+u^2}\\ 0\leq &u&\leq 1\end{eqnarray*} The bounds for \(u\) have been obtained in the following way. The bounds for \(x\) were \(0\) and \(\frac{\pi}2\). The corresponding bounds for \(\frac{x}2\) are \(0\) and \(\frac{\pi}4\), hence the bounds for \(u\) are \(\tan 0=0\) and \(\tan\frac{\pi}4=1\). The integral now becomes: \begin{eqnarray*} \int_0^{\frac{\pi}2}\frac1{10 -6\cos x}\,dx&=&\int_0^1\frac{\frac{2\,du}{1+u^2}}{10-6\frac{1-u^2}{1+u^2}}=\int_0^1\frac{2\,du}{(10-6)+(106)u^2}=\int_0^1 \frac{2\,du}{4 +16u^2}\\ &=&\frac1{16}\int_0^1\frac{2\,du}{\frac14+u^2}=\left.\frac2{8}\tan^{-1}\left(2u\right)\right|_0^1=\frac2{8}\tan^{-1}2. \end{eqnarray*} In the last line we used the formula \(\displaystyle \int\frac{1}{x^2+a^2}\,dx=\frac1a\tan^{-1}\frac{x}a+C\).