How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$

Question

I tried with multiplying with $\csc^{2}\left(x\right)$ and performing a $u$-substitution, but the roots got all complicated.

I really cannot see any other way to go about this.

Help would be highly appreciated.

Answer 1

Substitute $t=\sin x$ $$I= \int \frac1{t^{3/2}(1+t)\sqrt{1-t}}dt $$ Then, $1-t=y^2$ $$I= -\int \frac2{(1-y^2)^{3/2}}-\frac2{(2-y^2) \sqrt{1-y^2}} \ dy $$ As a result $$I= \frac{2y}{\sqrt{1-y^2}}-{\sqrt2}\tan^{-1}\frac{y}{\sqrt2\sqrt{1-y^2}}+C $$

Answer 2

By periodicity of the integrand and positivity of the argument of the radical, we may as well assume we are working on the interval $(0, \pi)$.

Since the integrand is an algebraic function in $\sin x$, applying the tangent half-angle substitution, $x = 2 \arctan t$, transforms the integral to an algebraic one: $$\frac1{\sqrt2} \int \frac{t^2 + 1}{t^{\frac32} (1 + t)} \,dt .$$ The form of this new integral in turn suggests the rationalizing substitution $t = u^2$, which gives $$\sqrt 2 \int \frac{1 + u^4}{u^2 (1 + u^2)} \,du = \sqrt 2 \int \left(1 - \frac2{1 + u^2} + \frac1{u^2}\right) \,du = \sqrt 2 \left(u - 2 \arctan u - \frac1u\right) + C .$$ Back-substituting yields $$\boxed{\int \frac{dx}{\sqrt{\sin^3 x + \sin^4 x}} = \sqrt {2 \tan \frac x2} - 2 \sqrt 2 \arctan \sqrt{\tan \frac x2} - \sqrt{2 \cot \frac x2} + C . }$$

Answer 3

The given integral is $$\int\frac{1}{\sqrt{\sin^{3}(x)+\sin^{4}(x)}}\mathrm dx$$

Trick:

Take $\sin^{4}(x)$ common from the square root in the denominator.

Therefore the integral will become $$\int\frac{1}{\sin^{2}(x)\sqrt{1+\csc(x)}}\mathrm dx$$

$$=\int\frac{\csc^{2}(x)}{\sqrt{1+\csc(x)}}\mathrm dx$$

$$=-2\int\frac{-\csc^{2}(x)\cot(x)}{2\sqrt{1+\csc(x)}\cot(x)}\mathrm dx$$

Now, substitute $\sqrt{1+\csc(x)}=u$

$$\implies \frac{-\csc(x)\cot(x)}{2\sqrt{1+\csc(x)}}\mathrm dx=\mathrm du$$

Now, try to understand this step.

$$\sqrt{1+\csc(x)}=u$$

$$\implies \csc(x)=u^{2}-1$$

$$\implies \csc^{2}(x)=(u^{2}-1)^{2}$$

$$\implies \cot^{2}(x)=(u^{2}-1)^{2}-1$$

$$\implies \cot^{2}(x)=u^{2}(u^{2}-2)$$

$$\implies \cot(x)=u\sqrt{u^{2}-2}$$

Therefore the integral is becoming

$$\int\frac{1}{\sqrt{\sin^{3}(x)+\sin^{4}(x)}}\mathrm dx =-2\int\frac{u^{2}-1}{u\sqrt{u^{2}-2}}\mathrm du$$

Now, try to evaluate $$-2\int\frac{u^{2}-1}{u\sqrt{u^{2}-2}}\mathrm du$$ using the concept of trigonometric substitution.

Substitute $u=\sqrt{2}\sec(\theta)$

Therefore $\mathrm du=\sqrt{2}×\sec(\theta)×\tan(\theta)\mathrm d\theta$

Therefore the integral will become

$$-2\int\frac{(2\sec^{2}(\theta)-1)(\sqrt{2}\sec(\theta)\tan(\theta))}{\sqrt{2}\sec(\theta)×\sqrt{2}\tan(\theta)}\mathrm d\theta$$

Finally the integral is becoming $$-2\int\frac{2\sec^{2}(\theta)-1}{\sqrt{2}}\mathrm d\theta$$

Now seperate and integrate it to get the result finally.

Best Wishes

Mathematics

Answer 4

Substitute $t=\sin x$:

$$\begin{align}\int\frac{\mathrm dx}{\sqrt{\sin^3x+\sin^4x}}&=\text{sgn}\sin2x\int\frac{\mathrm dt}{t^\frac32(1+t)\sqrt{1-t}}\\&=\text{sgn}\sin2x\int\frac{\mathrm dt}{(t^2+t)\sqrt{t-t^2}}\\&\overset{t=\frac1{u}}{=}-\text{sgn}\cos x\int\frac{u\mathrm du}{(u+1)\sqrt{u-1}}\\&=-\text{sgn}\cos x\int\frac{\mathrm du}{\sqrt{u-1}}+2\text{sgn}\cos x\int\frac{\mathrm d(\sqrt{u-1})}{\left(\sqrt{u-1}\right)^2+2}\\&=\text{sgn}(\cos x)\left(\sqrt2\tan^{-1}\sqrt{\frac{\csc x-1}2}-2\sqrt{\csc x-1}\right)+C\end{align}$$

Answer 5

Going by the first substitution that comes to mind, i.e., $t=\cot x$,

$$\begin{align}\mathcal I=\int\frac{\mathrm dx}{\sqrt{\sin^3x+\sin^4x}}&\overset{t=\cot x}{=}-\int\frac{\mathrm dt}{\sqrt{\text{sgn}(\csc x)\sqrt{t^2+1}+1}}\end{align}$$

Now, it seems like we have to make $2$ cases, but notice that the domain of the integrand is $\{x:\csc x>-1\}= \{x:\csc x\ge1\}\implies\text{sgn}(\csc x)=1$:

$$\begin{align}\mathcal I&=-\int\frac{\mathrm dt}{\sqrt{\sqrt{t^2+1}+1}}\end{align}$$

Whenever there are nested radicals like this in the integrand, it could be useful to check its inverse as in this question:

$$u=\sqrt{\sqrt{t^2+1}+1}$$ $$\implies t=\text{sgn}(t)u\sqrt{u^2-2}$$ $$\implies\mathrm dt=2\text{sgn}(t)\frac{u^2-1}{\sqrt{u^2-2}}\mathrm du$$

Hence,

$$\begin{align}\mathcal I&=-2 \text{sgn}(t)\int\frac{u^2-1}{u\sqrt{u^2-2}}\mathrm du\\&=2 \text{sgn}(t)\left(-\int\frac{u\mathrm du}{\sqrt{u^2-2}}+\int\frac{\mathrm du}{u\sqrt{u^2-2}}\right)\end{align}$$

Using the standard integral result

$$\int\frac{\mathrm dx}{|x|\sqrt{x^2-a^2}}=\frac1{a}\sec^{-1}\frac{x}{a}+C$$

the integral becomes $$\begin{align}\mathcal I&=2\text{sgn}(t)\left(-\sqrt{u^2-2}+ \sqrt2\sec^{-1}\frac{u}{\sqrt2}\right)+C\\&= \text{sgn}(\cot x)\left(-2\sqrt{\csc x-1}+\sqrt2\sec^{-1}\sqrt{\frac{\csc x+1}2}\right)+C\end{align}$$