0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
2
$\begingroup$

$$\int\frac{\sin^2x}{\cos^3x}dx=\int\frac{\cos{x}\sin^2x}{\cos^4x}dx=\int\frac{\sin^2x}{(1-\sin^2x)^2}d(\sin{x})=...$$ We can calculate the first integral by the same substitution.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

For $\int \frac{\sin^4 x}{\cos^3 x} dx$ try a smart part integration:

$$\int \frac{\sin^4 x}{\cos^3 x} dx = -\int \sin^3 x \frac{d\cos x}{\cos^3 x} = \frac{1}{2} \frac{\sin^3 x}{\cos^2 x} - \frac{3}{2} \int \frac{1-\cos^2 x}{\cos x} dx = ... $$

$\endgroup$
1
$\begingroup$

$$\int \frac { \sin ^{ 4 } x }{ \cos ^{ 3 } x } \, dx=\frac { 1 }{ 2 } \int { \sin ^{ 3 }{ x } d\left( \frac { 1 }{ \cos ^{ 2 }{ x } } \right) } =\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { \sin ^{ 2 }{ x } }{ \cos { x } } dx } \right) +C=\\ =\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { 1-\cos ^{ 2 }{ x } }{ \cos { x } } dx } \right) +C=\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { dx }{ \cos { x } } dx } +3\int { \cos { x } dx } \right) +C=\\ =\frac { 1 }{ 2 } \left( \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } -3\int { \frac { \cos { x } dx }{ \cos ^{ 2 }{ x } } dx } +3\int { \cos { x } dx } \right) +C=\frac { \sin ^{ 3 }{ x } }{ 2\cos ^{ 2 }{ x } } +\frac { 3\sin { x } }{ 2 } -\frac { 3 }{ 2 } \int { \frac { d\sin { x } }{ \left( 1-\sin { x } \right) \left( 1+\sin { x } \right) } +C } =\\ =\frac { \sin ^{ 3 }{ x } }{ 2\cos ^{ 2 }{ x } } +\frac { 3\sin { x } }{ 2 } -\frac { 3 }{ 4 } \left( \int { \frac { d\sin { x } }{ \left( 1-\sin { x } \right) } } +\int { \frac { d\sin { x } }{ \left( 1+\sin { x } \right) } } \right) +C=\frac { \sin ^{ 3 }{ x } }{ 2\cos ^{ 2 }{ x } } +\frac { 3\sin { x } }{ 2 } -\frac { 3 }{ 4 } \ln { \left| \frac { 1+\sin { x } }{ 1-\sin { x } } \right| } +C\\ \\ \\ \\ \\ $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.