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Exercise 9.3 · Q1

Q.Solve the following differential equation: dydx=1−cos⁡x1+cos⁡x\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

This is a direct integration problem after simplifying the right-hand side using trigonometric identities. The solution is y=2tan⁡x2−x+Cy = 2\tan\frac{x}{2} - x + C.

The equation is already in the form dydx=f(x)\frac{dy}{dx} = f(x), with no yy on the right. That means we don't need any special method like separation of variables — it's just pure integration. The challenge is purely algebraic: simplifying 1−cos⁡x1+cos⁡x\frac{1 - \cos x}{1 + \cos x} into something we can integrate easily.

Why use identities? Because direct integration of that ratio is messy. But if we rewrite it using half-angle formulas, the expression collapses into a clean sum of terms.

  1. Rewrite using half-angle identities. Recall: 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^2\frac{x}{2} 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2} So

1−cos⁡x1+cos⁡x=2sin⁡2x22cos⁡2x2=tan⁡2x2\frac{1 - \cos x}{1 + \cos x} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2}

  1. Express tan⁡2\tan^2 in integrable form. We know tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1. Therefore:

dydx=sec⁡2x2−1\frac{dy}{dx} = \sec^2\frac{x}{2} - 1

  1. Integrate both sides with respect to xx.

y=∫(sec⁡2x2−1)dxy = \int \left( \sec^2\frac{x}{2} - 1 \right) dx

For ∫sec⁡2x2 dx\int \sec^2\frac{x}{2}\, dx, let u=x2u = \frac{x}{2}, so dx=2 dudx = 2\, du, giving:

∫sec⁡2x2 dx=2∫sec⁡2u du=2tan⁡u=2tan⁡x2\int \sec^2\frac{x}{2}\, dx = 2\int \sec^2 u\, du = 2\tan u = 2\tan\frac{x}{2}

And ∫1 dx=x\int 1\, dx = x. So:

y=2tan⁡x2−x+Cy = 2\tan\frac{x}{2} - x + C

Watch out

A common mistake is forgetting the factor of 2 from the chain rule when integrating sec⁡2x2\sec^2\frac{x}{2}. Always check: derivative of tan⁡x2\tan\frac{x}{2} is 12sec⁡2x2\frac{1}{2}\sec^2\frac{x}{2}, so the integral must bring back a factor of 2.

Tip

You could also use the identity 1−cos⁡x1+cos⁡x=2sin⁡2x22cos⁡2x2=tan⁡2x2\frac{1-\cos x}{1+\cos x} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2} directly — no need to go through csc⁡\csc or cot⁡\cot forms. This is the cleanest path.

✓Final answer

The general solution is y=2tan⁡x2−x+Cy = 2\tan\frac{x}{2} - x + C, where CC is an arbitrary constant.

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