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Evaluate the indefinite integral $$\int\frac{\sin^4x}{\cos^3x}\,dx,~~\int\frac{\sin^2x}{\cos^3x}\,dx.$$
Attempt. We set $\tan(x/2)=t$, so: $$\cos x=\cos^2(x/2)-\sin^2(x/2)=\frac{1-t^2}{1+t^2},~~ \sin x=2\sin(x/2)\cos(x/2)=\frac{2t}{1+t^2}$$ and $\displaystyle dx=\frac{dt}{1+t^2},$ but this substitution leads us to complex calculations. Is there another, less complicated, approach?
Thank you!
$$\int\frac{\sin^2x}{\cos^3x}dx=\int\frac{\cos{x}\sin^2x}{\cos^4x}dx=\int\frac{\sin^2x}{(1-\sin^2x)^2}d(\sin{x})=...$$ We can calculate the first integral by the same substitution.
Use the trigonometric identity $\cos^{2}(x)+\sin^{2}(x)=1$ to transform these to \begin{align*} \int\frac{\sin^{4}(x)}{\cos^{3}(x)}\mathrm{d}x&=\int\frac{(1-\cos^{2}(x))^2}{\cos^{3}(x)}\mathrm{d}x\\&=\int\frac{1-2\cos^{2}(x)+\cos^{4}(x)}{\cos^{3}(x)}\mathrm{d}x\\ &=\int\sec^{3}(x)-2\sec(x)+\cos(x)\mathrm{d}x.\end{align*}
\begin{align*} \int\frac{\sin^{2}(x)}{\cos^{3}(x)}\mathrm{d}x&=\int\frac{1-\cos^{2}(x)}{\cos^{3}(x)}\mathrm{d}x\\&=\int\sec^{3}(x)-\sec(x)\mathrm{d}x. \end{align*}
You can find the integrals for $\cos,$ $\sec,$ and $\sec^{3}$ easily (for example, with Wolfram Alpha).
Bioche's rules lead to set $\;u=\sin x$, $\;\mathrm du=\cos x\,\mathrm d x$. The integral becomes the integral of a rational function: $$\int \frac{\sin^4x}{\cos^3 x}\,\mathrm d x=\int \frac{u^4}{\cos^3 x}\,\frac{\mathrm d u}{\cos x}=\int \frac{u^4}{(1-u^2)^2}\,\mathrm d u.$$
Let's decompose it into partial fractions: $$\frac{u^4}{(1-u^2)^2}=1+ \frac{2u^2-1}{(1-u^2)^2}=1+\frac{A}{1-u}+\frac{B}{(1-u)^2}+\frac{C}{1+u}+\frac{D}{(1+u)^2}.$$ As it is the decomposition of an even function, we see that $\;A=C$, $\;B=D$. Further, multiplying both sides by $(1-u^2)^2$, we obtain the identity: $$2u^2-1=A\bigl[(1-u)(1+u)^2+(1+u)(1-u)^2\bigr] + B\bigl[(1+u)^2+(1-u)^2\bigr],$$ so setting successively $u=1$, then $u=0$ yields $$B=D=\frac14, \quad 2A+2B=-1,\enspace\text{whence}\quad A=C=-\frac34.$$
Final computation: \begin{align} \int \frac{u^4}{(1-u^2)^2}\,\mathrm d u&=\int\mathrm d u-\frac34 \int\biggl(\frac{1}{1-u}+\frac{1}{1+u}\biggr)\mathrm d u+\frac14\int\biggl(\frac{1}{(1-u)^2}+\frac{1}{(1+u)^2}\biggr)\mathrm d u\\[1ex] &= u-\frac34 \ln\Bigl(\frac{1+u}{1-u}\Bigr)+\frac14\Bigl(\frac1{1-u}-\frac1{1+u}\Bigr)=u-\frac34 \ln\Bigl(\frac{1+u}{1-u}\Bigr)+\frac{u}{2(1-u^2)}\\[1ex] &=\sin x-\frac34 \ln\Bigl(\frac{1+\sin x}{1-\sin x}\Bigr)+\frac{\sin x}{2\cos^2x}. \end{align}
For $\int \frac{\sin^4 x}{\cos^3 x} dx$ try a smart part integration:
$$\int \frac{\sin^4 x}{\cos^3 x} dx = -\int \sin^3 x \frac{d\cos x}{\cos^3 x} = \frac{1}{2} \frac{\sin^3 x}{\cos^2 x} - \frac{3}{2} \int \frac{1-\cos^2 x}{\cos x} dx = ... $$
$$\int \frac { \sin ^{ 4 } x }{ \cos ^{ 3 } x } \, dx=\frac { 1 }{ 2 } \int { \sin ^{ 3 }{ x } d\left( \frac { 1 }{ \cos ^{ 2 }{ x } } \right) } =\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { \sin ^{ 2 }{ x } }{ \cos { x } } dx } \right) +C=\\ =\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { 1-\cos ^{ 2 }{ x } }{ \cos { x } } dx } \right) +C=\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { dx }{ \cos { x } } dx } +3\int { \cos { x } dx } \right) +C=\\ =\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { \cos { x } dx }{ \cos ^{ 2 }{ x } } dx } +3\int { \cos { x } dx } \right) +C=\frac { \sin ^{ 3 }{ x } }{ 2\cos ^{ 2 }{ x } } +\frac { 3\sin { x } }{ 2 } -\frac { 3 }{ 2 } \int { \frac { d\sin { x } }{ \left( 1-\sin { x } \right) \left( 1+\sin { x } \right) } +C } =\\ =\frac { \sin ^{ 3 }{ x } }{ 2\cos ^{ 2 }{ x } } +\frac { 3\sin { x } }{ 2 } -\frac { 3 }{ 4 } \left( \int { \frac { d\sin { x } }{ \left( 1-\sin { x } \right) } } +\int { \frac { d\sin { x } }{ \left( 1+\sin { x } \right) } } \right) +C=\frac { \sin ^{ 3 }{ x } }{ 2\cos ^{ 2 }{ x } } +\frac { 3\sin { x } }{ 2 } -\frac { 3 }{ 4 } \ln { \left| \frac { 1+\sin { x } }{ 1-\sin { x } } \right| } +C\\ \\ \\ \\ \\ $$
$$ \begin{aligned} \int \frac{\sin ^4 x}{\cos ^3 x} d x=&\frac{1}{2} \int \sin ^3 x d\left(\frac{1}{\cos ^2 x}\right) \\ \stackrel{IBP}{=} & \frac{1}{2} \frac{\sin ^3 x}{\cos ^2 x}-\frac{1}{2} \int \frac{1}{\cos ^2 x} \cdot 3 \sin ^2 x \cos x d x \\ =& \frac{\sin ^3 x}{2 \cos ^2 x}-\frac{3}{2} \int \frac{\sin ^2 x}{\cos x} d x \\ =& \frac{\sin ^3 x}{2 \cos ^2 x}-\frac{3}{2} \int \frac{1-\cos ^2 x}{\cos x} d x \\ =&\frac{\sin ^3 x}{2 \cos ^2 x}-\frac{3}{2} \int \sec x d x+\frac{3}{2}\int \cos x d x \\ =& \frac{\sin ^3 x}{2 \cos ^2 x}-\frac{3}{2} \ln |\sec x+\tan x|+\frac{3}{2} \sin x+C \end{aligned} $$
