I need to solve the integral below $$ \int\frac{dx}{\sin^2{x}\cos^3{x}} $$ without using hyperbolic functions but using substitutions like $u=\tan{x}$, $u=\sin{x}$ or $u=\cos{x}$.
Also, I know the correct answer: $$ \frac{3}{2}\ln\left | \tan{\frac{x}{2}+\frac{\pi}{4}} \right | - \frac{1}{\sin{x}}+\frac{\sin{x}}{2\cos^2{x}} $$ For me it looks like it will make sense to set $u=\sin{x}$ as long as $R(\sin{x}; -\cos{x}) = -R(\sin{x}; \cos{x})$. However, such substitution doesn't give me the expected result.


Note that\begin{align}\int\frac{\mathrm dx}{\sin^2x\cos^3x}&=\int\frac{\cos x}{\sin^2x\cos^4x}\,\mathrm dx\\&=\int\frac{\cos x}{\sin^2x(1-\sin^2x)^2}\,\mathrm dx.\end{align}So, by doing $\sin x=y$ and $\cos x\,\mathrm dx=\mathrm dy$, your indefinite integral becomes$$\int\frac{\mathrm dy}{y^2(1-y^2)^2}.$$Now, use the fact that\begin{align}\frac1{y^2(1-y^2)^2}&=\frac1{(1-y^2)^2}+\frac1{1-y^2}+\frac1{y^2}\\&=\frac3{4(y+1)}+\frac1{4(y+1)^2}-\frac3{4 (y-1)}+\frac1{4 (y-1)^2}+\frac1{y^2}.\end{align}


The given answer looks rather like an application of the substitution $t=\tan \frac x2$, which can be used directly after some rewriting of the integrand:

$$\frac{1}{\sin^2{x}\cos^3{x}} =\frac{\sin^2 x + \cos^2 x}{\sin^2{x}\cos^3{x}}=\frac 1{\cos^3x}+ \frac{1}{\sin^2x\cos x}$$

and again applying $1=\sin^2x + \cos^2x$ gives

$$\frac{1}{\sin^2{x}\cos^3{x}} = \frac{\sin^2 x}{\cos^3x} + \frac 2{\cos x} + \frac{\cos x}{\sin^2 x}$$

Now, we have

$$\int \frac{\cos x}{\sin^2 x} dx= -\frac 1{\sin x}(+C)$$


$$\int \sin x \frac{\sin x}{\cos^3x}dx = \frac{\sin x}{2\cos^2x}-\frac 12\int \frac 1{\cos x}dx$$

So, the integral becomes

$$\int \frac{1}{\sin^2{x}\cos^3{x}}dx = \frac{\sin x}{2\cos^2x} - \frac 1{\sin x}+\frac 32\int\frac 1{\cos x}dx$$

Handling $\int\frac 1{\cos x}dx$ by the substitution $t=\tan \frac x2$ gives

$$\int \frac 1{\cos x}dx= \ln \left\lvert\frac{1+\tan \frac x2}{1-\tan \frac x2}\right\rvert (+C)$$

Now, apply $\tan \frac{\pi}{4}=1$ and the addition formula for tangent

$$\frac{1+\tan \frac x2}{1-\tan \frac x2} = \frac{\tan \frac{\pi}{4}+\tan \frac x2}{1-\tan \frac{\pi}{4}\cdot\tan \frac x2} = \tan\left(\frac x2+\frac{\pi}{4}\right)$$

This gives the answer in the form you have it.


Why do you think that the substitution $u=\sin(x)$ doesn't give the expected result ?

If you don't edit your calculus one cannot check it.

$$ \int\frac{dx}{\sin^2{x}\cos^3{x}}=\int\frac{du}{u^2(1-u^2)^2}=-\frac{1}{u}+\frac{u}{2(1-u^2)}+\frac34\ln\left|\frac{u+1}{u-1}\right| $$ Then puting back $u=\sin(x)$ into it leads to the expected result.

  • $\begingroup$ there was a square missing ... just aded it $\endgroup$ Mar 10 '20 at 11:36
  • $\begingroup$ @trancelocation. OK. Thank you. $\endgroup$
    – JJacquelin
    Mar 10 '20 at 12:51
  • $\begingroup$ Yes. I learned this in calculus class: to integrate a power of $\sin x$ times a power of $\cos x$, where the exponent on $\cos x$ is odd, do this: keep one factor of $\cos x$ and convert all the others to $\sin x$. Here, the exponent on $\cos x$ is $-3$, which is odd. $\endgroup$
    – GEdgar
    Mar 11 '20 at 9:32

$$\dfrac1{\sin^2x\cos^3x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^3x}=\sec^3x+\dfrac1{\cos x\sin^2x}$$

For the first part, see this

For the last, set $\sin x=y,dy=?$ $$\int\dfrac1{\cos x\sin^2x}=\dfrac{\cos x}{(1-\sin^2x)\sin^2x}=\dfrac{dy}{(1-y^2)y^2}=\int\dfrac{dy}{y^2}+\int\dfrac{dy}{1-y^2}$$

  • $\begingroup$ For the fun of it, you should've applied the pythagorean identity once more :-) $\endgroup$ Mar 10 '20 at 11:59

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.