Definite Integration

Reduction Formulae

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Reduction Formulae

Reduction Formulae: Taming Definite Integrals

Hey JEE aspirants! Definite integration can seem daunting, especially when dealing with complex powers of trigonometric functions. That's where reduction formulae come to the rescue. They provide a systematic way to break down these integrals into simpler forms. Mastering these formulae is crucial for scoring well in JEE Main.

What are Reduction Formulae?

Imagine you have an integral like ∫sin⁡5x dx\int \sin^5 x \, dx. Calculating this directly can be tedious. A reduction formula allows you to express this integral, I5I_5, in terms of a simpler integral, say I3I_3 or I1I_1. It "reduces" the power of the trigonometric function, making the integration process much easier.

The core idea behind deriving reduction formulae is integration by parts. Remember that?

∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du

The trick is to cleverly choose uu and dvdv to achieve a reduction in the power of the trigonometric function within the integral.

Reduction Formula for ∫sin⁡nx dx\int \sin^n x \, dx

In=∫sin⁡nx dx=−sin⁡n−1x⋅cos⁡xn+n−1n⋅In−2I_n = \int \sin^n x \, dx = -\frac{\sin^{n-1}x \cdot \cos x}{n} + \frac{n-1}{n} \cdot I_{n-2}

Let's derive this! We have In=∫sin⁡nx dx=∫sin⁡n−1x⋅sin⁡x dxI_n = \int \sin^n x \, dx = \int \sin^{n-1} x \cdot \sin x \, dx.

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

Applying integration by parts:

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

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

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

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

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

In=−sin⁡n−1x⋅cos⁡x+(n−1)In−2−(n−1)InI_n = -\sin^{n-1} x \cdot \cos x + (n-1) I_{n-2} - (n-1) I_n

Rearranging to solve for InI_n:

In+(n−1)In=−sin⁡n−1x⋅cos⁡x+(n−1)In−2I_n + (n-1)I_n = -\sin^{n-1} x \cdot \cos x + (n-1) I_{n-2}

nIn=−sin⁡n−1x⋅cos⁡x+(n−1)In−2nI_n = -\sin^{n-1} x \cdot \cos x + (n-1) I_{n-2}

In=−sin⁡n−1x⋅cos⁡xn+n−1n⋅In−2I_n = -\frac{\sin^{n-1} x \cdot \cos x}{n} + \frac{n-1}{n} \cdot I_{n-2}

Voila! We have our reduction formula. Notice how InI_n is expressed in terms of In−2I_{n-2}. This means you can repeatedly apply this formula to reduce the power until you reach I1=∫sin⁡x dx=−cos⁡x+CI_1 = \int \sin x \, dx = -\cos x + C or I0=∫1 dx=x+CI_0 = \int 1 \, dx = x + C.

Reduction Formula for ∫cos⁡nx dx\int \cos^n x \, dx

In=∫cos⁡nx dx=cos⁡n−1x⋅sin⁡xn+n−1n⋅In−2I_n = \int \cos^n x \, dx = \frac{\cos^{n-1}x \cdot \sin x}{n} + \frac{n-1}{n} \cdot I_{n-2}

The derivation is very similar. We have In=∫cos⁡nx dx=∫cos⁡n−1x⋅cos⁡x dxI_n = \int \cos^n x \, dx = \int \cos^{n-1} x \cdot \cos x \, dx.

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

Applying integration by parts:

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

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

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

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

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

In=cos⁡n−1x⋅sin⁡x+(n−1)In−2−(n−1)InI_n = \cos^{n-1} x \cdot \sin x + (n-1) I_{n-2} - (n-1) I_n

Rearranging to solve for InI_n:

nIn=cos⁡n−1x⋅sin⁡x+(n−1)In−2nI_n = \cos^{n-1} x \cdot \sin x + (n-1) I_{n-2}

In=cos⁡n−1x⋅sin⁡xn+n−1n⋅In−2I_n = \frac{\cos^{n-1} x \cdot \sin x}{n} + \frac{n-1}{n} \cdot I_{n-2}

Again, we have a reduction formula expressing InI_n in terms of In−2I_{n-2}. Reduce until you reach I1=∫cos⁡x dx=sin⁡x+CI_1 = \int \cos x \, dx = \sin x + C or I0=∫1 dx=x+CI_0 = \int 1 \, dx = x + C.

Reduction Formula for ∫tan⁡nx dx\int \tan^n x \, dx

In=∫tan⁡nx dx=tan⁡n−1xn−1−In−2I_n = \int \tan^n x \, dx = \frac{\tan^{n-1}x}{n-1} - I_{n-2}

This one is slightly different. We rewrite tan⁡nx\tan^n x as tan⁡n−2x⋅tan⁡2x\tan^{n-2} x \cdot \tan^2 x.

So, In=∫tan⁡nx dx=∫tan⁡n−2x⋅tan⁡2x dxI_n = \int \tan^n x \, dx = \int \tan^{n-2} x \cdot \tan^2 x \, dx.

Using tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1:

In=∫tan⁡n−2x⋅(sec⁡2x−1) dxI_n = \int \tan^{n-2} x \cdot (\sec^2 x - 1) \, dx

In=∫tan⁡n−2x⋅sec⁡2x dx−∫tan⁡n−2x dxI_n = \int \tan^{n-2} x \cdot \sec^2 x \, dx - \int \tan^{n-2} x \, dx

The first integral is easy! The derivative of tan⁡x\tan x is sec⁡2x\sec^2 x, so:

∫tan⁡n−2x⋅sec⁡2x dx=tan⁡n−1xn−1+C\int \tan^{n-2} x \cdot \sec^2 x \, dx = \frac{\tan^{n-1} x}{n-1} + C

Therefore:

In=tan⁡n−1xn−1−In−2I_n = \frac{\tan^{n-1} x}{n-1} - I_{n-2}

Here, we reduce until we reach I1=∫tan⁡x dx=ln⁡∣sec⁡x∣+CI_1 = \int \tan x \, dx = \ln |\sec x| + C or I0=∫1 dx=x+CI_0 = \int 1 \, dx = x + C.

Tip: When applying reduction formulae in definite integrals, remember to change the limits of integration accordingly if you perform a substitution.
Common Mistake: Forgetting the constant of integration (CC) when applying reduction formulae in indefinite integrals. While it might not always be explicitly required in JEE, it's crucial to remember the general form.

JEE-Specific Trick: Often, JEE questions combine reduction formulae with other concepts like properties of definite integrals. Be prepared to identify the right approach by carefully observing the integrand and the limits.

Good luck, and happy integrating!