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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.

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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}

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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

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

and

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

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.

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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.

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

For the first part, see this

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

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

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