Reduction formula for ∫sinⁿ(x) dx — derive and apply for n=6

hard JEE-ADVANCED • JEE Advanced 2022 • 3 min read
Tags Integrals

Question

Derive the reduction formula for In=∫sin⁡nx dxI_n = \displaystyle\int \sin^n x\, dx. Then use it to evaluate ∫0π/2sin⁡6x dx\displaystyle\int_0^{\pi/2} \sin^6 x\, dx.

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

Write sin⁡nx=sin⁡n−1x⋅sin⁡x\sin^n x = \sin^{n-1} x \cdot \sin x.

Let u=sin⁡n−1xu = \sin^{n-1} x and dv=sin⁡x dxdv = \sin x\, dx. Then du=(n−1)sin⁡n−2xcos⁡x dxdu = (n-1)\sin^{n-2}x \cos x\, dx and v=−cos⁡xv = -\cos x.

 In=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2xcos⁡2x dxI_n = -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x \cos^2 x\, dx

Replace cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x:

In=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2x dx−(n−1)∫sin⁡nx dxI_n = -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x\, dx - (n-1)\int \sin^n x\, dx In=−sin⁡n−1xcos⁡x+(n−1)In−2−(n−1)InI_n = -\sin^{n-1}x \cos x + (n-1)I_{n-2} - (n-1)I_n In+(n−1)In=−sin⁡n−1xcos⁡x+(n−1)In−2I_n + (n-1)I_n = -\sin^{n-1}x \cos x + (n-1)I_{n-2} nIn=−sin⁡n−1xcos⁡x+(n−1)In−2nI_n = -\sin^{n-1}x \cos x + (n-1)I_{n-2} In=∫sin⁡nx dx=−sin⁡n−1xcos⁡xn+n−1nIn−2I_n = \int \sin^n x\, dx = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} I_{n-2}

For the definite integral ∫0π/2sin⁡nx dx\displaystyle\int_0^{\pi/2} \sin^n x\, dx, the boundary term vanishes (both sin⁡0=0\sin 0 = 0 and cos⁡(π/2)=0\cos(\pi/2) = 0), giving:

∫0π/2sin⁡nx dx=n−1n∫0π/2sin⁡n−2x dx\int_0^{\pi/2} \sin^n x\, dx = \frac{n-1}{n} \int_0^{\pi/2} \sin^{n-2} x\, dx

For n=6n = 6:

I6=56⋅I4=56⋅34⋅I2=56⋅34⋅12⋅I0I_6 = \frac{5}{6} \cdot I_4 = \frac{5}{6} \cdot \frac{3}{4} \cdot I_2 = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot I_0

Since I0=∫0π/21 dx=π2I_0 = \displaystyle\int_0^{\pi/2} 1\, dx = \frac{\pi}{2}:

I6=56⋅34⋅12⋅π2=5π32I_6 = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \mathbf{\frac{5\pi}{32}}

Why This Works

The reduction formula converts InI_n into In−2I_{n-2}, reducing the power by 2 each time. By repeatedly applying it, we reach either I0I_0 (if nn is even) or I1I_1 (if nn is odd), both of which are elementary integrals.

This is Wallis’ formula in action. The pattern for even nn:

∫0π/2sin⁡2kx dx=(2k−1)!!(2k)!!⋅π2\int_0^{\pi/2} \sin^{2k} x\, dx = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}

where !!!! denotes the double factorial.

Alternative Method

For sin⁡6x\sin^6 x specifically, you can use the power-reduction identity:

sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}

Cube both sides: sin⁡6x=18(1−cos⁡2x)3\sin^6 x = \frac{1}{8}(1 - \cos 2x)^3. Expand, integrate term by term. This avoids the reduction formula but involves more algebra.

For JEE Advanced, memorise the Wallis’ result: ∫0π/2sin⁡nx dx=∫0π/2cos⁡nx dx\int_0^{\pi/2} \sin^n x\, dx = \int_0^{\pi/2} \cos^n x\, dx, and the chain of fractions terminates at π/2\pi/2 for even nn and at 11 for odd nn. This saves time on definite integral problems.

Common Mistake

When deriving the reduction formula, students often forget to substitute cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x and instead try to integrate sin⁡n−2xcos⁡2x\sin^{n-2}x \cos^2 x directly. Without this substitution, you cannot bring InI_n back to the left side and the reduction does not close. The cos⁡2x→1−sin⁡2x\cos^2 x \to 1 - \sin^2 x step is the crucial trick.

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