Q.Integrate the function $\frac{1}{\sqrt{\sin^3 x\,\sin(x+\alpha)}}$

The key idea is to rewrite the product inside the square root using the sine addition formula, then simplify with a substitution.

First, expand $\sin(x+\alpha) = \sin x \cos \alpha + \cos x \sin \alpha$. The integrand becomes:

Factor $\sin^4 x$ out of the square root:

Now use the substitution $t = \cot x$, so $dt = -\csc^2 x \, dx = -\frac{1}{\sin^2 x} dx$. The integral becomes:

This is a standard power integral. Integrate:

Substitute back $t = \cot x$: …

Why U Substitution Works Here

When you see an integrand with $\sin^3 x$ and $\sin(x+\alpha)$, your first instinct might be to expand $\sin(x+\alpha)$ using the addition formula. That’s exactly what we need — but the trick is to notice that the product $\sin^3 x \cdot \sin(x+\alpha)$ can be rewritten in a way that reveals a hidden derivative.

The expression $\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$ looks messy, but if we factor out $\sin^4 x$ from the product inside the square root, we get something like $\sin^2 x \sqrt{\sin(x+\alpha)/\sin x}$. That ratio $\frac{\sin(x+\alpha)}{\sin x}$ simplifies to $\cos\alpha + \cot x \sin\alpha$, which is a linear function of $\cot x$. And the derivative of $\cot x$ is $-\csc^2 x$, which appears naturally when we manipulate the integrand.

So the plan is: rewrite the integrand so that it becomes a function of $\cot x$ times $-\csc^2 x$, then substitute $t = \cot x$.

Step-by-Step Solution

1. Rewrite the integrand using the sine addition formula

Therefore,

2. Factor $\sin^4 x$ out of the square root

So the integrand becomes:

3. Express in terms of $\cot x$ Recall that $\csc^2 x = 1 + \cot^2 x$, but more importantly, $\frac{1}{\sin^2 x} = \csc^2 x$. So:

4. Set up the substitution Let $t = \cot x$. Then $dt = -\csc^2 x \, dx$, so $\csc^2 x \, dx = -dt$. The integrand becomes:

5. Integrate with respect to $t$ The integral is now:

This is a standard power integral. Let $u = \cos\alpha + t \sin\alpha$, then $du = \sin\alpha \, dt$, so $dt = \frac{du}{\sin\alpha}$. But it’s easier to directly integrate:

where $A = \cos\alpha$ and $B = \sin\alpha$. Thus: …

Method: Factor $\sin^{4}x$ out of the radical to expose a $\cot x$ substitution

Use this for integrands like $\dfrac{1}{\sqrt{\sin^{3}x\,\sin(x+\alpha)}}$: expanding the shifted sine and factoring a power of $\sin x$ turns everything into a function of $\cot x$ times $\csc^2 x$.

Steps

Step 1: Expand the shifted sine.

$\sin(x+\alpha)=\sin x\cos\alpha+\cos x\sin\alpha$, so $\sin^3x\sin(x+\alpha)=\sin^4x(\cos\alpha+\cot x\,\sin\alpha)$.

Step 2: Pull $\sin^4x$ out of the root. …

Common Mistakes

Mistake 1: Not factoring $\sin^4x$ out of the radical.

Why it's wrong: without writing $\sqrt{\sin^3x\sin(x+\alpha)}=\sin^2x\sqrt{\cos\alpha+\cot x\sin\alpha}$, the $\cot x$ substitution never appears. Correct approach: expand $\sin(x+\alpha)$ and factor $\sin^4x$.

Mistake 2: Dropping the minus from $d(\cot x)=-\csc^2x\,dx$.

Why it's wrong: it flips the sign of the final result. Correct approach: substitute $\csc^2x\,dx=-dt$. …