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Exercise 7.3 · Q9

Q.Integrate the following function: cos⁡x1+cos⁡x\frac{\cos x}{1 + \cos x}

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The key idea is to rewrite the integrand using the identity cos⁡x=2cos⁡2(x/2)−1\cos x = 2\cos^2(x/2) - 1, then simplify to a form that integrates directly. The result is x−tan⁡(x/2)+Cx - \tan(x/2) + C.

Why This Approach Works

When you see a rational function of cos⁡x\cos x, the standard trick is to use the half-angle substitution t=tan⁡(x/2)t = \tan(x/2). That works, but it can get messy. Here, there's a cleaner path: rewrite cos⁡x\cos x in terms of cos⁡(x/2)\cos(x/2) using the double-angle identity. This turns the denominator into something that cancels nicely, leaving you with a simple sum of two terms — one constant, one a standard trigonometric integral.

The intuition: the denominator 1+cos⁡x1 + \cos x is exactly 2cos⁡2(x/2)2\cos^2(x/2). That's the key simplification. Once you see that, the rest is straightforward.

Step-by-Step Solution

  1. Rewrite the denominator using the half-angle identity. Recall that cos⁡x=2cos⁡2(x/2)−1\cos x = 2\cos^2(x/2) - 1. Therefore:

1+cos⁡x=1+(2cos⁡2(x/2)−1)=2cos⁡2(x/2).1 + \cos x = 1 + (2\cos^2(x/2) - 1) = 2\cos^2(x/2).

This is the crucial simplification — the denominator becomes a perfect square of a cosine.

  1. Rewrite the numerator in terms of cos⁡(x/2)\cos(x/2) as well. The numerator is cos⁡x\cos x. Using the same identity:

cos⁡x=2cos⁡2(x/2)−1.\cos x = 2\cos^2(x/2) - 1.

So the integrand becomes:

cos⁡x1+cos⁡x=2cos⁡2(x/2)−12cos⁡2(x/2).\frac{\cos x}{1 + \cos x} = \frac{2\cos^2(x/2) - 1}{2\cos^2(x/2)}.

  1. Split the fraction into two simpler terms. Divide each term in the numerator by the denominator:

2cos⁡2(x/2)2cos⁡2(x/2)−12cos⁡2(x/2)=1−12sec⁡2(x/2).\frac{2\cos^2(x/2)}{2\cos^2(x/2)} - \frac{1}{2\cos^2(x/2)} = 1 - \frac{1}{2}\sec^2(x/2).

Now the integrand is expressed as a constant minus a constant times sec⁡2\sec^2 of half the angle.

  1. Integrate term by term. The integral becomes: ∫(1−12sec⁡2(x/2))dx=∫1 dx−12∫sec⁡2(x/2) dx.\int \left(1 - \frac{1}{2}\sec^2(x/2)\right) dx = \int 1\,dx - \frac{1}{2}\int \sec^2(x/2)\,dx. …

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